Question

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

Answer 1

`\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.

Step-by-step 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})} = (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^(-)}] \\ &= (n)/(V) \approx (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}`.