# Suppose 10.00mL of a hydrochloric acid solution with a pH of 1.097 is titrated with a sodium hydroxide with a pH of 13.00. How many mL of the sodium hydroxide solution will be required to neutralize the acid?

May 31, 2018

The titration will require 8.00 mL of $\text{NaOH}$.

#### Explanation:

The equation for the reaction is

$\text{H"_3"O"^"+" + "OH"^"-" → "2H"_2"O}$

1. Calculate the concentration of $\text{H"_3"O"^"+}$

$\text{pH = 1.097}$

["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-1.097"color(white)(l)"mol/L" = "0.0800 mol/L" = "0.0800 mmol/mL"

2. Calculate the moles of $\text{H"_3"O"^"+}$

$\text{Moles of H"_3"O"^"+" = 10.00 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) ×( "0.0800 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "0.800 mmol H"_3"O"^"+}$

3. Calculate the moles of $\text{OH"^"-}$ needed

$\text{Moles of OH"^"-" = 0.800 color(red)(cancel(color(black)("mmol H"_3"O"^"+"))) × ("1 mmol OH"^"-")/(1 color(red)(cancel(color(black)("mmol H"_3"O"^"+")))) = "0.800 mmol OH"^"-}$

4. Calculate the concentration of $\text{OH"^"-}$

$\text{pH = 13.00}$

$\text{pOH = 14.0 - pH = 14.00 - 13 00 = 1.00}$

["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-1.00"color(white)(l)"mol/L" = "0.100 mol/L" = "0.100 mmol/mL"

5. Calculate the volume of the $\text{OH"^"-}$

$\text{Volume of OH"^"-" = 0.800 color(red)(cancel(color(black)("mmol OH"^"-"))) × ("1 mL OH"^"-")/(0.100 color(red)(cancel(color(black)("mmol OH"^"-")))) = "8.00 mL OH"^"-}$

The titration will require 8.00 mL of $\text{NaOH}$.