Question

We find that 18.90 milliliters of a 2.50M KOH solution are required to titrate 35.0 milliters of a H2S04 solution. What is the molarity of the H2S04 solution? 2 2KOH + H2S04 K2S04 2 H20 A. 0.675 M B. 1.57 M C. 0.0711 M D. 0.128 M E. 4.04 M F. 0.903 M G. 1.18 M H. 0.0173 M

Answer 1

Answer:

Concentration of $H_{2}SO_{4}$ is 0.675 M

Explanation:

According to balanced equation, 2 moles of KOH neutralizes 1 mol of $H_{2}SO_{4}$.

18.90 mL of 2.50M KOH = $\frac{2.50\times 18.90}{1000}moles$ of KOH = 0.04725 moles of KOH

If molarity of $H_{2}SO_{4}$ is C (M) then-

moles of $H_{2}SO_{4}$ are neutralized =  $\frac{C\times 35.0}{1000}moles$ of $H_{2}SO_{4}$

Hence, $\frac{1}{2}\times 0.04725=\frac{C\times 35.0}{1000}$

or, C = 0.675

So, concentration of $H_{2}SO_{4}$ is 0.675 M