Question

what volume, in milliliters, of 0.100 m hclo₃ is required to neutralize 46.1 ml of 0.140 m koh?
answer:

ml

Explanation:

Step1: Write the neutralization reaction

$$\text{HClO}_3+\text{KOH}=\text{KClO}_3+\text{H}_2\text{O}$$
From the reaction, the mole ratio of $\text{HClO}_3$ to $\text{KOH}$ is $1:1$.

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

Use the formula $n = C\times V$ (where $n$ is moles, $C$ is concentration, and $V$ is volume in liters).
$V_{\text{KOH}}=46.1\ \text{mL}=0.0461\ \text{L}$, $C_{\text{KOH}} = 0.140\ \text{M}$
$n_{\text{KOH}}=C_{\text{KOH}}\times V_{\text{KOH}}=0.140\times0.0461 = 0.006454\ \text{mol}$

Step3: Calculate the volume of $\text{HClO}_3$

Since $n_{\text{HClO}_3}=n_{\text{KOH}}$ (mole ratio $1:1$), and $C_{\text{HClO}_3}=0.100\ \text{M}$
Using $V=\frac{n}{C}$, $V_{\text{HClO}_3}=\frac{n_{\text{HClO}_3}}{C_{\text{HClO}_3}}=\frac{0.006454}{0.100}=0.06454\ \text{L}$
Convert liters to milliliters: $V_{\text{HClO}_3}=0.06454\times1000 = 64.5\ \text{mL}$

Answer:

$64.5$