Question #bdb47

1 Answer
Nov 28, 2016

Answer:

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

Explanation:

You have the following species in solution.

From #"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 #"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 #"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 #"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 #"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 #"H"_3"O"^+# from the #"H"_2"SO"_4# will neutralize the total of 0.0100 mol of #"OH"^"-"# from the #"NaOH"# and #"Ba(OH)"_2#.

That leaves the reaction between #"KOH"# and #"HOCl"#.

#"HOCl"# is a weak acid. It is partially neutralised by the #"KOH"#.

We have generated a buffer.

#color(white)(mmmmml)"HOCl" color(white)(ll)+ color(white)(m)"OH"^"-"color(white)(ll) → color(white)(m)"OCl"^"-" + "H"_2"O"#
#"I/mol:"color(white)(mll)"0.003 00"color(white)(ml)"0.001 50"color(white)(mmmm)0#
#"C/mol:"color(white)(m)"-0.001 50"color(white)(m)"-0.001 50"color(white)(m)"+0.001 50"#
#"E/mol:"color(white)(ml)"0.001 50"color(white)(mmml)0 color(white)(mmmll)"0.001 50"#

For #"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#