Question

Calculate the poH if 2.95 g of KOH(s) dissolved to make 100mL of solution

Answer 1

Answer:

$\rm pOH \approx 1.279$ if $2.95\; \rm g$ of $\rm KOH$ is dissolved completely in water to make a $100\; \rm mL$ solution.

Explanation:

Look up the relative atomic mass data on a modern periodic table:

• $\rm K$: $39.098$.
• $\rm O$: $15.999$.
• $\rm H$: $1.008$.

Calculate the formula mass of $\rm KOH$:

$\begin{aligned}&M(\mathrm{KOH})\\ &= 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}$.

Find the number of moles of formula units in that $2.95\; \rm g$ of $\rm KOH$:

$\begin{aligned}& n(\mathrm{KOH}) \\ & = \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} = \frac{2.95\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.0525800\; \rm mol \end{aligned}$.

$\rm KOH$ is a strong base. When dissolved in water, it ionizes completely to produce $\rm K^{+}$ ions and hydroxide ions ($\rm OH^{-}$.)

Note that there are one moles of $\rm OH^{-}$ ions in each mole of $\rm KOH$ formula units. When that $2.95\; \rm g$ of $\rm KOH$ (approximately $0.0525800\; \rm mol$ of $\rm KOH$ formula units) dissolves completely in water to make a $100\; \rm mL$ solution, about $0.0525800\; \rm mol$ of $\rm OH^{-}$ ions will be produced.

Convert the unit of volume to liters: $V = 100\; \rm mL = 0.1\; \rm L$.

Calculate the concentration of $\rm OH^{-}$ ions in that solution:

$\begin{aligned}&[\mathrm{OH^{-}}] \\ &= \frac{n}{V} \approx \frac{0.0525800\; \rm mol}{0.1 \; \rm L} \approx 0.525800\; \rm mol \cdot L^{-1} \end{aligned}$.

Calculate the $\rm pOH$ of that solution:

$\begin{aligned}& \mathrm{pOH} \\ &= -\log_{10}[\mathrm{OH^{-}}] \\ &\approx -\log_{10}(0.525800) \approx 1.279\end{aligned}$.