Skip to main content
Chemistry

Return to Question

Fixed \ce usage in preparation of upcoming MathJax/mhchem update
Source Link
edit approved Feb 16, 2016 at 23:10
mhchem
edit approved Feb 16, 2016 at 23:10
mhchem
  • 3.3k
  • 2
  • 17
  • 35

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align} \ce{ M^{3+} + {H2O} &~<=> {M(OH)^{2+}} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ {M(OH)^{2+}} + {H2O} &~<=> {M(OH)^{+}_{2}} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ {M(OH)^{+}_{2}} + {H2O} &~<=> {M(OH)_3} + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}\begin{align} \ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align} \ce{ M^{3+} + {H2O} &~<=> {M(OH)^{2+}} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ {M(OH)^{2+}} + {H2O} &~<=> {M(OH)^{+}_{2}} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ {M(OH)^{+}_{2}} + {H2O} &~<=> {M(OH)_3} + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align} \ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Notice removed Draw attention by CommunityBot
Bounty Ended with no winning answer by CommunityBot
inserted alignment characters
Source Link
Martin - マーチン ♦
Martin - マーチン ♦
  • 44.4k
  • 13
  • 161
  • 327

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}3} = 9.0 \end{align} $$ All\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad \mathrm{p}K_{\mathrm{a}3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$\begin{align} \ce{ M^{3+} + {H2O} &~<=> {M(OH)^{2+}} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ {M(OH)^{2+}} + {H2O} &~<=> {M(OH)^{+}_{2}} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ {M(OH)^{+}_{2}} + {H2O} &~<=> {M(OH)_3} + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $(b)$$\mathrm{(b)}$ and $(c)$$\mathrm{(c)}$ it can be assumed that reaction $(a)$$\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}3} = 9.0 \end{align} $$ All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad \mathrm{p}K_{\mathrm{a}3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $(b)$ and $(c)$ it can be assumed that reaction $(a)$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + {OH^{-}} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align} \ce{ M^{3+} + {H2O} &~<=> {M(OH)^{2+}} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ {M(OH)^{2+}} + {H2O} &~<=> {M(OH)^{+}_{2}} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ {M(OH)^{+}_{2}} + {H2O} &~<=> {M(OH)_3} + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

removed buzzword and improved formatting
Source Link
ringo
ringo
  • 24.1k
  • 6
  • 89
  • 136

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, 0.1M$0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad pK_{b1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad pK_{b2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad pK_{b3} = 9.0 \end{align} $$$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}3} = 9.0 \end{align} $$ All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad pK_{a1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad pK_{a2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad pK_{a3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad \mathrm{p}K_{\mathrm{a}3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $pK_a$$\mathrm{p}K_\mathrm{a}$ values of $(b)$ and $(c)$ it can be assumed that reaction $(a)$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Thanks a lot.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, 0.1M $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad pK_{b1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad pK_{b2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad pK_{b3} = 9.0 \end{align} $$ All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad pK_{a1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad pK_{a2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad pK_{a3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $pK_a$ values of $(b)$ and $(c)$ it can be assumed that reaction $(a)$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Thanks a lot.

Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

$$ \begin{align} \ce{ M(OH)3 <=> M(OH)^{+}_{2} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}1} = 0.5 \\ \ce{ M(OH)^{+}_{2} <=> M(OH)^{2+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}2} = 0.7 \\ \ce{ M(OH)^{2+} <=> M^{3+} + {OH^{-}} } \qquad \mathrm{p}K_{\mathrm{b}3} = 9.0 \end{align} $$ All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

$$\begin{align} \ce{ M^{3+} + {H2O} <=> {M(OH)^{2+}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}1} = 14.0 - 9.0 = 5.0 \quad ---(a) \\ \ce{ {M(OH)^{2+}} + {H2O} <=> {M(OH)^{+}_{2}} + H+ } \qquad \mathrm{p}K_{\mathrm{a}2} = 14.0 - 0.7 = 13.3 \quad ---(b)\\ \ce{ {M(OH)^{+}_{2}} + {H2O} <=> {M(OH)_3} + H+ } \qquad \mathrm{p}K_{\mathrm{a}3} = 14.0 - 0.5 = 13.5 \quad ---(c) \end{align} $$

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $(b)$ and $(c)$ it can be assumed that reaction $(a)$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Tweeted twitter.com/#!/StackChemistry/status/608296183929434112
Notice added Draw attention by user16096
Bounty Started worth 50 reputation by CommunityBot
Source Link
user16096
user16096