# What is the pH of a solution in which "25.0 mL" of a "0.100-M" solution of "NaOH" has been added to "100. mL" of a "0.100-M" "HCl" solution?

Oct 18, 2017

$\text{pH} = 1.222$

#### Explanation:

As you know, sodium hydroxide and hydrochloric acid neutralize each other in a $1 : 1$ mole ratio as described by the balanced chemical equation

${\text{NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

This means that a complete neutralization, which would result in a neutral solution, i.e. a solution that has $\text{pH} = 7$ at room temperature, requires equal numbers of moles of sodium hydroxide and hydrochloric acid.

Notice that your two solutions have equal molarities, but that the volume of the hydrochloric acid solution is

$\left(100. \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))))/(25.0color(red)(cancel(color(black)("mL}}}}\right) = 4$

times larger than the volume of the sodium hydroxide solution. This implies that the number of moles of hydrochloric acid is $4$ times bigger than the number of moles of sodium hydroxide.

This means that after the reaction is complete, you will be left with excess hydrochloric acid $\to$ the $\text{pH}$ of the resulting solution will be $< 7$.

Now, the number of moles of hydrochloric acid that will not take part in the reaction is given by

overbrace(100. color(red)(cancel(color(black)("mL"))) * "0.100 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))))^(color(blue)("moles of HCl added")) - overbrace(25.0 color(red)(cancel(color(black)("mL"))) * "0.100 moles NaOH"/(10^3color(red)(cancel(color(black)("mL")))))^(color(blue)("moles of NaOH added"))

 = underbrace(((75 * 0.100)/10^3)color(white)(.)"moles HCl")_(color(blue)("what is not consumed by the reaction"))

The total volume of the resulting solution will be

$\text{25.0 mL + 100. mL = 125 mL}$

As you know, the $\text{pH}$ of the solution is given by

"pH" = - log(["H"_3"O"^(+)])

Since hydrochloric acid is a strong acid that ionizes in a $1 : 1$ mole ratio to produce hydronium cations, you can say that the concentration of hydronium cations in the resulting solution will be

["H"_3"O"^(+)] = (((75 * 0.100)/color(blue)(cancel(color(black)(10^3))))color(white)(.)"moles H"_3"O"^(+))/(125 * color(blue)(cancel(color(black)(10^3)))color(white)(.)"L") = ((75 * 0.100)/125)"mol L"^(-1)

This means that you have

"pH" = - log((75 * 0.100)/125) = color(darkgreen)(ul(color(black)(1.222)

The answer is rounded to three decimal places, the number of sig figs you have for your values.