Question

how many milliliters of a 0.309 m hcl solution are needed to completely neutralize 233 ml of a 0.0295 m ba(oh)₂ solution?

Explanation:

Step1: Write the balanced chemical equation

$$\text{Ba(OH)}_2 + 2\text{HCl}=\text{BaCl}_2 + 2\text{H}_2\text{O}$$
From the equation, the mole ratio of $\text{Ba(OH)}_2$ to $\text{HCl}$ is $1:2$.

Step2: Calculate the moles of $\text{Ba(OH)}_2$

Use the formula $n = C\times V$ (where $n$ is moles, $C$ is concentration, and $V$ is volume in liters).
$V_{\text{Ba(OH)}_2}=233\space mL = 0.233\space L$, $C_{\text{Ba(OH)}_2}=0.0295\space M$
$n_{\text{Ba(OH)}_2}=0.0295\space M\times0.233\space L = 0.0068735\space mol$

Step3: Calculate the moles of $\text{HCl}$

Since the mole ratio of $\text{Ba(OH)}_2$ to $\text{HCl}$ is $1:2$, $n_{\text{HCl}} = 2\times n_{\text{Ba(OH)}_2}$
$n_{\text{HCl}}=2\times0.0068735\space mol= 0.013747\space mol$

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

Use the formula $V=\frac{n}{C}$ (where $V$ is volume in liters, $n$ is moles, $C$ is concentration).
$C_{\text{HCl}} = 0.309\space M$, $n_{\text{HCl}}=0.013747\space mol$
$V_{\text{HCl}}=\frac{0.013747\space mol}{0.309\space M}\approx0.0445\space L$

Convert liters to milliliters: $V_{\text{HCl}}=0.0445\space L\times1000\space mL/L = 44.5\space mL$

Answer:

$44.5\space mL$