Question

what volume, in milliliters, of 0.20 m hcl is required to neutralize 30.0 ml of 0.80 m naoh? answer: ml

Explanation:

Step1: Write the neutralization reaction

$$\text{HCl}+\text{NaOH}=\text{NaCl}+\text{H}_2\text{O}$$
From the reaction, the mole ratio of $\text{HCl}$ to $\text{NaOH}$ is $1:1$.

Step2: Calculate the moles of $\text{NaOH}$

Use the formula $n = M\times V$ (where $n$ is moles, $M$ is molarity, and $V$ is volume in liters).
For $\text{NaOH}$: $M = 0.80\ M$, $V=30.0\ mL=0.0300\ L$
$n_{\text{NaOH}}=0.80\ M\times0.0300\ L = 0.024\ mol$

Step3: Find moles of $\text{HCl}$

Since mole ratio $\text{HCl}:\text{NaOH}=1:1$, $n_{\text{HCl}} = n_{\text{NaOH}}=0.024\ mol$

Step4: Calculate volume of $\text{HCl}$

For $\text{HCl}$: $M = 0.20\ M$, $n = 0.024\ mol$
Using $V=\frac{n}{M}$, $V_{\text{HCl}}=\frac{0.024\ mol}{0.20\ M}=0.12\ L$

Step5: Convert volume to milliliters

$0.12\ L\times1000\ mL/L = 120\ mL$

Answer:

$120$