Solution
- The reation of the acid-base reaction is given below:
[tex]HNO_3+KOH\to KNO_3+H_2O[/tex]
- The number of moles of the base and acid are 1 each from the equation above.
- That is,
[tex]\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}[/tex]
- Next, we can simply apply the formula below to find the molarity of the base KOH.
[tex]\begin{gathered} \frac{C_aV_a}{C_bV_b}=\frac{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}[/tex]
- Thus, we can find the molarity of KOH as follows:
[tex]\begin{gathered} \frac{0.75\times25}{C_b\times15}=\frac{1}{1} \\ \\ \text{ Multiply both sides by }C_b \\ \\ C_b=\frac{0.75\times25}{15} \\ \\ C_b=1.25M \end{gathered}[/tex]
Final Answer
The answer is 1.25M
