1. Calculation of EMF

## TASK 4

## SOLUTION

The amount of potassium hydroxide and nitric acid in the solutions:

$$
\begin{gathered}
v(\mathrm{KOH})=\omega \rho V / M=0.062 \cdot 1.055 \text { g/ml } \cdot 22.7 \text { ml / } 56 \text { g/ mol }=0.0265 \text { mol. } \\
v\left(\mathrm{HNO}_{3}\right)=\text { C }_{\mathrm{M}} \cdot \mathrm{V}=2.00 \text { mol/l } \cdot 0.0463 \text { l }=0.0926 \text { mol. }
\end{gathered}
$$

Since $v(\mathrm{KOH})<v(\mathrm{HNO} 3)$, and the coefficients of these substances in the reaction equation are the same and equal to one, potassium hydroxide is in deficiency, and its amount will determine the heat effect of the reaction, which will be:

$$
\text { Q = 55.6 kJ/mol } \cdot \text { 0.0265 mol = 1.47 kJ. }
$$
Answer: 1.47 kJ of heat will be released.

## Grading system: