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Topic: Stuck on a pH calculation Is my logic faulty  (Read 1393 times)

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Offline JLA

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Stuck on a pH calculation Is my logic faulty
« on: May 26, 2018, 05:39:37 AM »
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

Offline Borek

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Re: Stuck on a pH calculation Is my logic faulty
« Reply #1 on: May 26, 2018, 11:12:58 AM »
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.
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Offline JLA

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Re: Stuck on a pH calculation Is my logic faulty
« Reply #2 on: May 27, 2018, 08:45:12 AM »
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.

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