Question

A volume of 50.0 mL of aqueous potassium hydroxide ( KOH ) was titrated against a standard solution of sulfuric acid ( H2SO4 ). What was the molarity of the KOH solution if 25.7 mL of 1.50 M H2SO4 was needed? The equation is2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answer 1

Answer

The molarity of the KOH = 1.542 M

Explanation

The given parameters are:

Volume of KOH, Vb = 50.0 mL

Volume of H2SO4, Va = 25.7 mL

Molarity of H2SO4, Ca = 1.50 M

Equation: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

From the equation, the mole ratio of KOH to H2SO4 = 2:1, that is na = 1 and nb = 2

What to find:

The molarity of the KOH, Cb.

Step-by-step solution:

The molarity of the KOH, Cb can be calculated using the formula below:

$\begin{gathered} \frac{C_aV_a}{n_a}=\frac{C_bV_b}{n_b} \\ \\ C_b=\frac{C_a\times V_a\times n_b}{n_a\times V_b} \end{gathered}$

Substituting the values of the parameters into the formula, we have

$\begin{gathered} C_b=\frac{1.50M\times25.7mL\times2}{1\times50.0mL} \\ \\ C_b=\frac{77.1M}{50.0} \\ \\ C_b=1.542\text{ }M \end{gathered}$

Thus, the molarity of the KOH is 1.542 M