Question

an aqueous solution of perchloric acid is standardized by titration with a 0.194 m solution of potassium hydroxide. if 11.9 ml of base are required to neutralize 29.5 ml of the acid, what is the molarity of the perchloric acid solution?

Explanation:

Step1: Write the neutralization reaction

Perchloric acid ($HClO_4$) reacts with potassium hydroxide ($KOH$) in a 1:1 molar - ratio: $HClO_4 + KOH=KClO_4 + H_2O$. The formula for titration is $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of the acid, and $M_2$ and $V_2$ are the molarity and volume of the base.

Step2: Identify the given values

We are given that $V_1 = 29.5\ mL$ (volume of perchloric acid), $M_2=0.194\ M$ (molarity of $KOH$), and $V_2 = 11.9\ mL$ (volume of $KOH$). We need to find $M_1$.

Step3: Rearrange the titration formula to solve for $M_1$

From $M_1V_1 = M_2V_2$, we can solve for $M_1$ as $M_1=\frac{M_2V_2}{V_1}$.

Step4: Substitute the values into the formula

$M_1=\frac{0.194\ M\times11.9\ mL}{29.5\ mL}$.
First, calculate the numerator: $0.194\ M\times11.9\ mL = 2.3086\ mmol$ (millimoles).
Then, divide by the denominator: $M_1=\frac{2.3086\ mmol}{29.5\ mL}= 0.0783\ M$.

Answer:

$0.0783\ M$