[JLA]

Q) Next, the student decided to mix together solutions of sodium hydroxide, NaOH, and barium hydroxide, Ba(OH)₂.
The original sodium hydroxide solution had a pH of 9.9 and the student measured 850 mL of this into a beaker. He then added 95 mL of a 0.075 mol L-1 barium hydroxide solution to the same beaker.  Calculate the pH of the resulting solution.  (8 marks)

I keep trying to make this calculation workout using two approaches. The problem is I get two different answers!  Can you see if or why one of the approaches should NOT work  OR if you can make approach 2 give pH 12 please show me how?

1)NaOH given pH=9.9, V=0.850 L: Then [H⁺]=1.2589x10^-10 mol/L.  Therefore [OH^-]=Kw/1.2589x10^-10=7.94328x10^-5 mol/L.  Therefore moles OH^- = 7.94328x10^-5 * 0.850 = 6.75179x10^-5mol.

Ba(OH)₂ given c=0.075 mol/L, V=0.095 L:  Then moles OH^- = 0.075*0.092*2 = 0.01425 mol.

total moles OH^- = 6.75179x10^-5 + 0.01425 = 0.0143175 moles.  Total concentration of OH^- = 0.0143175/0.945 = 0.0151508 mol/L.   Therefore total [H⁺] = Kw/0.0151508 = 6.6003x10^-13 mol/L

pH=-log(6.6003x10^-13)=12.18

2)  NaOH given pH=9.9, V=0.850 L:  Then [H⁺]=1.2589x10^-10 mol/L.  Therefore moles H⁺ = c*V = 1.2589x10^-10 mol/L * 0.850 L = 1.0700866x10^-10 moles.

Ba(OH)₂ given c=0.075 mol/L, V=0.095 L: Then [OH-]=2*0.075 mol/L = 0.15 mol/L.  Therefore [H⁺]=Kw/0.15 = 6.6667x10^-14 mol/L. Therefore moles H⁺ = c*V = 6.6667x10^-14 mol/L * 0.095 L = 6.3333x10^-15 mole.

total moles H⁺ = 1.0700866x10^-10 + 6.3333x10^-15 = 1.0701x10^-10 moles
total concentration H⁺ = total n/total V = 1.0701x10^-10/0.945 = 1.1324x10^-10 mol/L.

pH=-log(1.1324x10^-10) = 9.9

If you can see why my second approach won't work please tell me!  Thanks, Jenny

[Borek]

First answer is correct.

Most of the H⁺ from the NaOH solution will react with added excess of OH^- from Ba(OH)₂ solution. Actually this reaction will also consume some OH^-, but removing 10^-10 from 0.075 M doesn't change the number by too much.

[JLA]

Thank you.  Yes, I see now that the second approach will not work as the hydrogen from the NaOH will be reacted with the large excess of the concentrated Ba(OH)₂.