Answer
There are necessary 0.16 g of potassium hydroxide $(KOH)$, to prepare a solution with pH = 11.56
Work Step by Step
1. Calculate the hydroxide ion concentration:
pH + pOH = 14
11.56 + pOH = 14
pOH = 2.44
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 2.44}$
$[OH^-] = 3.6 \times 10^{- 3}M$
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- Since $KOH$ is a strong base: $[KOH]$ = $[OH^-]$.
- Molar mass (KOH): 39.10* 1 + 16.00* 1 + 1.008* 1 = $56.11g/mol$
- Solution volume = 800.0mL
2. Use these informations to calculate the necessary mass of $KOH$ to prepare a solution with pH = 11.56
$3.6 \times 10^{-3}M (OH^-) \times \frac{1mol(KOH)}{1mol(OH^-)} \times \frac{800.0mL}{1} \times \frac{1L}{1000mL} \times \frac{56.11g(KOH)}{1mol(KOH)} = 0.16g(KOH)$
