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?
1 Answer
Explanation:
As you know, sodium hydroxide and hydrochloric acid neutralize each other in a
#"NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#
This means that a complete neutralization, which would result in a neutral solution, i.e. a solution that has
Notice that your two solutions have equal molarities, but that the volume of the hydrochloric acid solution is
#(100. color(red)(cancel(color(black)("mL"))))/(25.0color(red)(cancel(color(black)("mL")))) = 4#
times larger than the volume of the sodium hydroxide solution. This implies that the number of moles of hydrochloric acid is
This means that after the reaction is complete, you will be left with excess hydrochloric acid
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
#"25.0 mL + 100. mL = 125 mL"#
As you know, the
#"pH" = - log(["H"_3"O"^(+)])#
Since hydrochloric acid is a strong acid that ionizes in a
#["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.