Question

What is the pH of a solution made by mixing 100.0 mL of 0.10 M `HNO_3`, 50.0 mL of 0.20 M `HCl` and 100.0 mL of water? Assume that the volumes are additive.

Answer 1

`"pH" = 1.10`

Explanation:

First thing first, calculate the total volume of the resulting solution

`V_"total" = "100.0 mL + 50.0 mL + 100.0 mL"`

`V_"total" = "250.0 mL"`

Now, you are dealing with two strong acids that ionize completely in aqueous solution. Both nitric acid and hydrochloric acid produce hydronium cations in `1:1` mole ratios, so you know that

`["H"_ 3"O"^(+)]_ ("coming from HNO"_ 3) = ["HNO"_3]`

`["H"_ 3"O"^(+)]_ ("coming from HCl") = ["HCl"]`

As you know, molarity is defined as the number of moles of solute present in `10^3` `"mL"` of solution. For the nitric acid solution, you have

`"100.0 mL" = (10^3color(white)(.)"mL")/color(blue)(10) => n_( "H"_3"O"^(+)) = "0.10 moles"/color(blue)(10) = "0.010 moles H"_3"O"^(+)`

For hydrochloric acid, you have

`"50.0 mL"= (10^3color(white)(.)"mL")/color(blue)(20) implies n_ ("H"_ 3"O"^(+)) = "0.20 moles"/color(blue)(20) = "0.010 moles H"_3"O"^(+)`

The total number of moles of hydronium cations delivered by the two acids in the resulting solution will be

`n_ ("H"_ 3"O"^(+)) = "0.010 moles + 0.010 moles"`

`n_ ("H"_ 3"O"^(+)) = "0.020 moles"`

The concentration of the hydronium cations in the resulting solution will be

`["H"_3"O"^(+)] = "0.020 moles"/(250.0 * 10^(-3)"L") = "0.080 M"`

As you know, you have

`color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))`

This will give you

`color(darkgreen)(ul(color(black)("pH" = - log(0.080) = 1.10)))`

The answer is rounded to two decimal places, the number of sig figs you have for the molarities of the two acids.

As a fun fact, a mixture of nitric acid and hydrochloric acid is called aqua regia. Ideally, aqua regia contains nitric acid and hydrochloric acid in a `1:3` mole ratio, not in a `1:1` mole ratio like you have here.