Answer
$[OH^-] = 0.1128M$
$pH = 13.052$
Work Step by Step
1. Calculate the molar mass:
Molar Mass ($KOH$):
39.1* 1 + 16* 1 + 1.01* 1 = 56.11g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 3.165}{ 56.11}$
$n(moles) = 0.05641$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
- 500ml = 0.5L
$ C(mol/L) = \frac{ 0.05641}{ 0.5} $
$C(mol/L) = 0.1128$
- Since KOH is a strong base:$[OH^-] = [KOH] = 0.1128M$
4. Now, calculate the pOH, and then the pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.1128)$
$pOH = 0.943$
$pH + pOH = 14$
$pH + 0.9476 = 14$
$pH = 13.052$