# Question b7b14

Mar 28, 2017

The pH is 1.64.

#### Explanation:

The balanced equation is

$\text{HNO"_3 + "KOH" → "KNO"_3 + "H"_2"O}$

or

$\text{H"^"+" + "OH"^"-" → "H"_2"O}$

${\text{Initial moles of H"^+ = "0.040 00" color(red)(cancel(color(black)("L HNO"_3))) × (0.100 color(red)(cancel(color(black)("mol HNO"_3))))/(1 color(red)(cancel(color(black)("L HNO"_3)))) × ("1 mol H"^+)/(1 color(red)(cancel(color(black)("mol HNO"_3)))) = "0.004 00 mol H}}^{+}$

$\text{Moles of OH"^"-" color(white)(l)"added" = "0.025 00" color(red)(cancel(color(black)("L KOH"))) × (0.100 color(red)(cancel(color(black)("mol KOH"))))/(1 color(red)(cancel(color(black)("L KOH")))) × ("1 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.002 50 mol OH"^"-}$

You have more moles of $\text{H"^"+}$ than of $\text{OH"^"-}$, so all of the $\text{OH"^"-}$ will be neutralized.

The moles of $\text{H"^"+}$ remaining are

$\text{(0.004 00 - 0.002 50) mol" = "0.001 50 mol}$

The total volume is now

$\text{(40.00 + 25.00) mL = 65.00 mL = 0.065 00 L}$

["H"^"+"] = "moles"/"litres" = ("0.001 50 mol")/("0.065 00 L") = "0.023 08 mol/L"#

$\text{pH" = "-log"["H"^"+"] = "-log(0.023 08)} = 1.64$