Question

A solution of potassium permanganate $\left({\mathrm{KMnO}}_4\right)$ was found by titration to be 0.05138M at$\mathbf{24}\mathbf{°}\mathbf{C}$ . What is the molarity when the lab temperature drops to$\mathbf{16}\mathbf{°}\mathit{C}$ ?

Short answer

Thus, the concentration of potassium permanganate ( ${\mathrm{KMnO}}_4$) at $16°C$ is 0.05146M.

Step-by-step solution

Discussing the correction for thermal expansion

The concentration (molarity) of the solution changes as the temperature changes, and this concentration change at a given temperature is estimated using a mathematical procedure called the correction for thermal expansion.

$\frac{C\text{'}}{d}=\frac{c}{d}$,

where,

$C\text{'}$is the concentration of the solution at the applied temperature,

role="math" localid="1663310790523" $d\text{'}$is the density of the solution at the applied temperature,

c is the concentration of the solution at the applied temperature, and

d is the density of the solution at the applied temperature.

Calculating the concentration of potassium permanganate ( KMnO4) at 16°C

Given:

The concentration of ${\mathrm{KMnO}}_4$at $24°C$is 0.05138M .

The density of water at $24°C$is 0.997299g/mL .

The density of water at $16°C$is 0.997299g/mL .

The correction for thermal expansion is

$\frac{c\text{'}}{d}=\frac{c}{d}$.

The concentration and density of water at $16°C$and $24°C$are plugged into the above equation, and some rearrangement is performed to produce the concentration of potassium permanganate ( $KMn{O}_4$) at $16°C$.

$\frac{\mathrm{c}\text{'}\mathrm{at}16°\mathrm{c}}{0.9989460\mathrm{g}/\mathrm{mL}}=\frac{0.05138\mathrm{M}}{0.9972995\mathrm{g}/\mathrm{mL}}$

$C\text{'}$at$16°\mathrm{C}=\frac{0.05138\mathrm{M}}{0.9972995\mathrm{g}/\mathrm{mL}}\times 0.9989460\mathrm{g}/\mathrm{mL}$

$=0.05146\mathrm{M}$.

Thus, the concentration of potassium permanganate ( ${\mathrm{KMnO}}_4$) at $16°C$is $0.05146\mathrm{M}$.