Question

35.0 mL of stock hydrochloric acid solution is added to a 500 mL volumetric flask. A student adds distilled water up to the line. Using a pH meter and logarithms, he finds the molarity of the new solution to be 0.062 M HCl. What is the molarity of the stock solution?

Answer 1

Answer:

Molarity of stock HCl solution is 0.89 M

Explanation:

The given problem can be solved using laws of dilution.

According to laws of dilution-    $C_{1}V_{1}=C_{2}V_{2}$

Where, $C_{1}$ and $C_{2}$ are initial and final concentration of a solution

            $V_{1}$ and $V_{2}$ are initial and final volume of a solution

Here, $V_{1}=35.0mL$, $C_{2}=0.062M$ and $V_{2}=500mL$

So, $C_{1}=\frac{C_{2}V_{2}}{V_{1}}$ = $\frac{(0.062M)\times (500mL)}{35.0mL}$ = 0.89 M

Hence, molarity of stock HCl solution is 0.89 M