Answer
$[OH^-] \approx 0.3758M$
$pH \approx 13.575$
Work Step by Step
1. Find the molar mass of the base:
Li: 6.94 * 1
O: 16 * 1
H: 1.01 * 1
6.94 + 16 + 1.01 = 23.95 g/mol
2. Find the nº of moles in 2.250g of $LiOH$:
$mm = mass(g) \div n(moles)$
$23.95 = 2.250 \div n(moles)$
$n(moles) = 2.250 \div 23.95$
$n(moles) = 0.09395$
3. Calculate the concentration of $LiOH$:
$C = n(moles) \div V(L)$
$C = 0.09395 \div 0.250$
$C = 0.3758M$
4. Since one LiOH is a strong base, we will consider $[LiOH] = [OH^-]$
$[OH^-] = 0.3758M$
5. Find the pOH and them the pH:
$pOH = -log[OH^-]$
$pOH = -log(0.3758)$
$pOH \approx 0.425 $
$pH = 14 - pOH$
$pH = 14 - 0.425$
$pH \approx 13.575$
