Question

what volume of a 0.234 m hydrochloric acid solution is required to neutralize 11.6 ml of a 0.103 m sodium hydroxide solution?
use the references to access important values if needed for this question.
submit answer
retry entire group
9 more group attempts remaining

Explanation:

Step1: Write the neutralization reaction equation

The reaction between hydrochloric acid ($HCl$) and sodium hydroxide ($NaOH$) is $HCl + NaOH=NaCl + H_2O$. The mole ratio of $HCl$ to $NaOH$ is $1:1$.

Step2: Calculate the number of moles of $NaOH$

Use the formula $n = C\times V$, where $C$ is the concentration and $V$ is the volume. For $NaOH$, $C = 0.103\ M$ and $V=11.6\ mL=11.6\times10^{- 3}\ L$. So $n_{NaOH}=0.103\ M\times11.6\times10^{-3}\ L = 0.103\times11.6\times10^{-3}\ mol$.
$$n_{NaOH}=1.1948\times10^{-3}\ mol$$

Step3: Calculate the volume of $HCl$

Since $n_{HCl}=n_{NaOH}$ (from the mole - ratio in the reaction), and for $HCl$, $C = 0.234\ M$. Using the formula $V=\frac{n}{C}$, we substitute $n = 1.1948\times10^{-3}\ mol$ and $C = 0.234\ M$.
$$V_{HCl}=\frac{1.1948\times 10^{-3}\ mol}{0.234\ mol/L}$$
$$V_{HCl}=5.106\times10^{-3}\ L$$
Convert liters to milliliters: $V_{HCl}=5.106\times10^{-3}\ L\times1000\ mL/L = 5.11\ mL$

Answer:

$5.11\ mL$