# Question 8dd3e

Apr 13, 2017

The pH will be 12.2

#### Explanation:

First, we calculate the amount of solute in each solution, using moles of solute = conc. of solution x volume (in L)

HI: (0.10 mol/L)(0.210 L) = 0.0210 mol

NaOH: (0.10 mol/L)(0.100 L) = 0.0100 mol

LiOH: (0.30 mol/L)(0.0550 L) = 0.0165 mol

Since all of these are strong electrolytes, each will dissociate completely in its solution. Also, HI is monoprotic (producing only one hydronium ion), so the total amount of hydronium and hydroxide ions will be:

${H}_{3} {O}^{+}$ : 0.021 mol (from the HI).

$O {H}^{-}$ : 0.0265 mol (adding the amounts from the two bases together).

Neutralization is 1 : 1, so the hydronium ions react with an equal number of hydroxide ions:

${H}_{3} {O}^{+} + O {H}^{-}$ $\rightarrow 2 {H}_{2} O$

and 0.0055 mol of $O {H}^{-}$ remains in a total volume of 0.365 L (adding all three volumes together).

Therefore $\left[O {H}^{-}\right] = \frac{0.0055 \text{mol}}{0.365 L} = 0.0151$ mol/L

to find pOH, take -log(0.0151) = 1.8

This means the pH is 14 - 1.8 = 12.2