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

The molarity of the KOH = 1.542 M

Step-by-step 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} (C_aV_a)/(n_a)=(C_bV_b)/(n_b) \\ \\ C_b=(C_a* V_a* n_b)/(n_a* V_b) \end{gathered}`

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

`\begin{gathered} C_b=(1.50M*25.7mL*2)/(1*50.0mL) \\ \\ C_b=(77.1M)/(50.0) \\ \\ C_b=1.542\text{ }M \end{gathered}`

Thus, the molarity of the KOH is 1.542 M