A workbook question asks:

$\pu{2.184g}$ of a solid mixture containing only $\ce{K2CO3}$ ($\mathrm{FW = 138.2058}\pu{ g/mol}$) and $\ce{KHCO3}$ ($\mathrm{FW = 100.1154} \pu{g/mol}$) is dissolved in distilled water. $\pu{31.29mL}$ of a $\pu{0.742\!M}\ \ce{HCl}$ standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of $\ce{K2CO3}$ and $\ce{KHCO3}$ in the mixture.

I started this problem by finding the amount of moles of $\ce{HCl}$ added to the titration, assuming that the solution was completely neutralized at the endpoint.

In $\pu{0.03129L}$ of a solution of $\pu{0.742\!M}\ \ce{HCl}$, there are $$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Then I set up the balanced equations:

$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$ $$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

This should indicate that the $\ce{HCl}$ will combine with the carbonate and bicarbonate in a 2:1 ratio, favoring the carbonate, since two moles of HCl are required to neutralize one mole of carbonate.


$$\ce{\pu{0.0232mol}\ HCl \times \frac{1\; mol\; CO_3}{2\; mol\; HCl} = 0.0116\; mol\; CO3}$$

$$\ce{0.0232\; mol\; HCl \times \frac{1\; mol\; HCO_3}{1 \pu{mol} HCl} = 0.0232\; mol\; HCO3} $$


$$\ce{0.0116\; mol\; CO3 \times \frac{1\; mol\; K2CO3}{1\; mol\; CO3} = 0.0116\; mol\; K2CO3}$$

$$\ce{0.0232\; mol\; HCO3 \times \frac{1\; mol\; KHCO3}{1\; mol\; HCO3} = \pu{0.0232\; mol}\; KHCO3}$$

Converting moles to grams:

$$ \ce{0.0116\; mol\; K2CO3 \times \frac{138.2058\; g}{mol\; K2CO3} = 1.60\; g} \\ \ce{0.0232\; mol\; KHCO3 \times \frac{100.1154\; g}{mol\; KHCO3} = 2.32\; g} $$

When I add these masses together, I get $\pu{3.92 g}$, which is greater than $\pu{2.184 g}$.

Where did I go wrong?

  • 3
    HINT - This is linear algebra with two equations and two unknowns. So there are two equations relating moles(K2CO3) and moles(KHCO3). – MaxW Sep 14 at 13:23
  • 2
    Welcome to Chemistry.SE Please note some of the edits made. It will make your life easier notably \$\ce{}\$ for chemical equations and formulas, \$\mathrm{}\$ for mathmatical equations and \$\pu{}\$ for units. – A.K. Sep 14 at 14:29
up vote 1 down vote accepted

We first start by writing the system of equations in this reaction, assuming that we don't know the ratios that the elements combine in.

Because we know the mass of the unknown mixture, we'll use $x$ to represent the mass of $\ce{K2CO3}$ and $y$ to represent the mass of $\ce{KHCO3}$, in grams. This is our first equation: $$x + y = 2.184$$

Next, we know the final amount of moles of $\ce{HCl}$ that reacted. It was $$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Now, let's reconsider the balanced equations for the reaction. We have $$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$ $$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

We can see that it takes $\ce{2 HCl}$ to neutralize one $\ce{CO3}$, but it only takes $\ce{1 HCl}$ to neutralize one $\ce{HCO3}$.

We also know the formula weights of the individual components. Now, we can write the second equation as an expression of the sum of the moles of the reactants. Our coefficients for $x$ and $y$ become: $$ \ce{\frac{2\; mol\; HCl}{1\; mol\; K2CO3} \times \frac{1\; mol\; K2CO3}{138.2058\; g}} \times x $$

$$ \ce{\frac{1\; mol\; HCl}{1\; mol KHCO3} \times \frac{1\; mol\; KHCO3}{100.1154\; g}} \times y $$

Now we can cancel out units and write out the second equation: $$\ce{\frac{2\; mol\; HCl}{138.2058\; g}}x + \ce{\frac{1\; mol\; HCl}{100.1154\; g}}y = \ce{0.0232\; mol\; HCl}$$

After that, this is straightforward linear algebra. This system of equations is simple enough to be solved by basic substitution and elimination. However, you can choose to use a matrix. In that case, we have

$$ \mathrm{ \begin{bmatrix} 1 & 1 \\ \frac{2}{138.2058} & \frac{1}{100.1154} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} } = \begin{bmatrix} 2.184 \\ 0.0232 \end{bmatrix} $$

Rearranging gives $$ \mathrm{ \begin{bmatrix} 1 & 1 \\ \frac{2}{138.2058} & \frac{1}{100.1154} \end{bmatrix}^{-1} \begin{bmatrix} 2.184 \\ 0.0232 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} } $$

Using a calculator we find that $$\mathrm{ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0.309\\ 1.875 \end{bmatrix} } $$ $$x = \ce{0.309\; g\; K2CO3}$$ $$y = \ce{1.875\; g\; KHCO3}$$

Finally, the mass percentages are determined to be: $$\ce{\frac{0.309\; g\; K2CO3}{2.184\; g\; total} \times 100{\%} = 14.1{\%} K2CO3}$$ $$\ce{\frac{1.875\; g\; KHCO3}{2.184\; g\; total} \times 100{\%} = 85.9{\%} KHCO3} $$

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