in a titration, 10.0 ml of 0.0750 m hcl(aq) is exactly neutralized by 30.0 ml of koh(aq) of unknown…

in a titration, 10.0 ml of 0.0750 m hcl(aq) is exactly neutralized by 30.0 ml of koh(aq) of unknown concentration. what is the concentration of the koh(aq) solution?\na) 0.0250 m\nb) 0.0750 m\nc) 0.225 m\nd) 0.333 m

in a titration, 10.0 ml of 0.0750 m hcl(aq) is exactly neutralized by 30.0 ml of koh(aq) of unknown concentration. what is the concentration of the koh(aq) solution?\na) 0.0250 m\nb) 0.0750 m\nc) 0.225 m\nd) 0.333 m

Answer

Explanation:

Step1: Write the neutralization reaction

$HCl(aq)+KOH(aq)\rightarrow KCl(aq) + H_2O(l)$; the mole - ratio of $HCl$ to $KOH$ is $1:1$.

Step2: Calculate the moles of $HCl$

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume in liters. $V_{HCl}=10.0\ mL = 0.0100\ L$ and $M_{HCl}=0.0750\ M$. So $n_{HCl}=M_{HCl}\times V_{HCl}=0.0750\ mol/L\times0.0100\ L = 7.50\times 10^{-4}\ mol$.

Step3: Determine the moles of $KOH$

Since the mole - ratio of $HCl$ to $KOH$ is $1:1$, $n_{KOH}=n_{HCl}=7.50\times 10^{-4}\ mol$.

Step4: Calculate the molarity of $KOH$

$V_{KOH}=30.0\ mL = 0.0300\ L$. Use the formula $M=\frac{n}{V}$, so $M_{KOH}=\frac{n_{KOH}}{V_{KOH}}=\frac{7.50\times 10^{-4}\ mol}{0.0300\ L}=0.0250\ M$.

Answer:

A. $0.0250\ M$