# Question bdb47

##### 1 Answer
Nov 28, 2016

WARNING! Long answer! The pH of the solution is 7.53.

#### Explanation:

You have the following species in solution.

From ${\text{H"_2"SO}}_{4}$:

0.0500 color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.100 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × (2 "mol H"_3"O"^+)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.0100 mol H"_3"O"^+

From $\text{NaOH}$:

0.0250 color(red)(cancel(color(black)("L NaOH"))) × (0.200 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × (1 "mol OH"^"-")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.005 00 mol OH"^"-"

From ${\text{Ba(OH)}}_{2}$:

0.0250 color(red)(cancel(color(black)("L Ba(OH)"_2))) × (0.100 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × ("2 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol Ba(OH)"_2)))) = "0.005 00 mol OH"^"-"

From $\text{KOH}$:

0.0100 color(red)(cancel(color(black)("L KOH"))) × (0.150 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")))) = "0.001 50 mol OH"^"-"

From $\text{HOCl}$:

0.0300 color(red)(cancel(color(black)("L HOCl"))) × (0.100 "mol HOCl")/(1 color(red)(cancel(color(black)("L HOCl")))) = "0.003 00 mol HOCl"

The 0.0100 mol of ${\text{H"_3"O}}^{+}$ from the ${\text{H"_2"SO}}_{4}$ will neutralize the total of 0.0100 mol of $\text{OH"^"-}$ from the $\text{NaOH}$ and ${\text{Ba(OH)}}_{2}$.

That leaves the reaction between $\text{KOH}$ and $\text{HOCl}$.

$\text{HOCl}$ is a weak acid. It is partially neutralised by the $\text{KOH}$.

We have generated a buffer.

$\textcolor{w h i t e}{m m m m m l} \text{HOCl" color(white)(ll)+ color(white)(m)"OH"^"-"color(white)(ll) → color(white)(m)"OCl"^"-" + "H"_2"O}$
$\text{I/mol:"color(white)(mll)"0.003 00"color(white)(ml)"0.001 50} \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol:"color(white)(m)"-0.001 50"color(white)(m)"-0.001 50"color(white)(m)"+0.001 50}$
$\text{E/mol:"color(white)(ml)"0.001 50"color(white)(mmml)0 color(white)(mmmll)"0.001 50}$

For $\text{HOCl", "p"K_"a} = 7.53$.

According to the Henderson-Hasselbalch equation

"pH" = "p"K_"a" + log(("[OCl"^"-""]")/"[HOCl]") = 7.53 + log((stackrelcolor(blue)(1)(color(red)(cancel(color(black)("0.001 50"color(white)(l) "mol/L")))))/(color(red)(cancel(color(black)("0.001 50" color(white)(l)"mol/L"))))) = 7.53#