# Titration, finding the molarity of HCl

$25~\mathrm{mL}$ of $\ce{HCl}$ are titrated with $.104~\mathrm{M}$ $\ce{NaOH}$ solution. If $32.55~\mathrm{mL}$ of $\ce{NaOH}$ are required to standardize, what is the molarity of $\ce{HCl}$? This is what I have so far:

1. How I got the molarity of $\ce{NaOH}$:

• $.8234~\mathrm{g}$ KHP requires $38.76~\mathrm{mL}$ of $\ce{NaOH}$ to standardize
• $.8234~\mathrm{g} \times 1~\mathrm{mol} \times (201.22~\mathrm{g})^{-1} = 4.032\times 10^{-3}~\mathrm{mol}~\ce{NaOH}$
• $4.032\times 10^{-3} \times (.03876~\mathrm{L})^{-1} = .104~\mathrm{M}~\ce{NaOH}$ solution.
2. $.025~\mathrm{L} \times x = .03255~\mathrm{L} \times .104$
$x = .1354~\mathrm{M}~\ce{HCl}$

I think that this is correct but I want to be sure.

• Let me start by saying I improved the formatting of your post with MathJax. For more information on how to do so yourself, check out the help center and this meta post. Give me a second to think over the calculation; it seems to me you are doing it in a too complicated way. – Jan Nov 1 '15 at 19:26
• Yes, that is the correct answer but there is a much simpler method to do this question – Nanoputian Nov 1 '15 at 19:30

Yes, your answer is correct, but let me show you a simpler way to arrive there. First, remember that the equation of the acid-base reaction is as follows:

$$\ce{NaOH + HCl -> NaCl + H2O}$$

Meaning that we use $\ce{NaOH}$ and $\ce{HCl}$ in a $1:1$ ratio. Therefore we know:

$$n(\ce{NaOH}) = n(\ce{HCl})$$

At the equivalence point. Remember that concentrations are defined as:

$$c = \frac{n}{V} \Longrightarrow n = c V$$

$$n(\ce{HCl}) = n(\ce{NaOH}) \\ c(\ce{HCl}) \cdot V(\ce{HCl}) = c(\ce{NaOH}) \cdot V(\ce{NaOH}) \\ c(\ce{HCl}) = \frac{c(\ce{NaOH}) \cdot V(\ce{NaOH})}{V(\ce{HCl})} \\ c(\ce{HCl}) = \frac{0.104~\mathrm{M} \times 32.55~\mathrm{ml}}{25~\mathrm{ml}} \\ c(\ce{HCl}) = 0.135~\mathrm{M}$$