# What is the concentration of hydrochloric acid solution if 6.68 ml is neutralised by 13.90 ml of 0.0161 M potassium hydroxide solution ?

Jun 14, 2015

The molarity of the hydrochloric acid solution was 0.0335 M.

#### Explanation:

You're dealing with a neutralization reaction in which hydrochloric acid, a strong acid, reacts with potassium hydroxide, a strong base, to produce potassium chloride, a salt, and water.

The balanced chemical equation looks like this

$H C {l}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \to K C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

The important thing to notice here is that you have a $1 : 1$ mole ratio between hydrochloric acid and potassium hydroxide. This means that, in order to get a complete neutralization, you need equal numbers of moles of acid and of base.

Since molarity is defined as moles of solute per liters of solution, you can use the volume and molarity of the potassium hydroxide solution to determine how many moles of base were needed

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{K O H} = \text{0.0161 M" * 13.90 * 10^(-3)"L" = "0.0002238 moles}$ $K O H$

This means that your hydrochloric acid solution contained

${n}_{H C l} = {n}_{K O H} = \text{0.0002238 moles}$ $H C l$

Now use the volume of the hydrochloric solution and the number of moles of $H C l$ it contains to determine its molarity

C_(HCl) = n_(HCl)/V_(HCl) = "0.0002238 moles"/(6.68 * 10^(-3)"L")= color(green)("0.0335 M")

The answer is rounded to three sig figs, the number of sig figs you gave for the molarity of the potassium hydroxide solution.

Jun 15, 2015

The concentration of HCl = $0.033 \text{mol/l}$

#### Explanation:

$H C {l}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \rightarrow K C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

$c = \frac{n}{v}$

$n = c v$

So no. moles KOH = $\frac{13.90 \times 0.0161}{1000} = 2.23 \times {10}^{- 4}$

From the equation we can see that they react in a molar ratio of 1:1.

This means that the no. moles HCl must be the same:

$n H C l = 2.23 \times {10}^{- 4}$

Since $c = \frac{n}{v}$:

$\left[H C l\right] = \frac{2.23 \times {10}^{- 4}}{6.68 \times {10}^{- 3}} = 0.033 \text{mol/l}$