Question

Neutralization occurs when 15.0 mL of KOH reacts with 25.0 mL of HNOg. If the molarity ofHNOg is 0.750 M, what is the molarity of the KOH?

Answer 1

- The reation of the acid-base reaction is given below:

`HNO_3+KOH\to KNO_3+H_2O`

- The number of moles of the base and acid are 1 each from the equation above.

- That is,

`\begin{gathered} n_a=1 \\ n_b=1 \\ \text{ where,} \\ n_a,n_b\text{ are the number of moles of acid and base in the reaction} \end{gathered}`

- Next, we can simply apply the formula below to find the molarity of the base KOH.

`\begin{gathered} (C_aV_a)/(C_bV_b)=(n_a)/(n_b) \\ \\ where, \\ C_a,C_b\text{ are the concentrations of acid and base} \\ V_a,V_b\text{ are the volumes of the acid and base} \\ \\ \\ \text{ We have been given:} \\ C_a=0.75M \\ C_b=? \\ V_a=25mL \\ V_b=15mL \end{gathered}`

- Thus, we can find the molarity of KOH as follows:

`\begin{gathered} (0.75*25)/(C_b*15)=(1)/(1) \\ \\ \text{ Multiply both sides by }C_b \\ \\ C_b=(0.75*25)/(15) \\ \\ C_b=1.25M \end{gathered}`

Final Answer

The answer is 1.25M