I tried solving this question, and I know the answer since it was given, but I c
ID: 780365 • Letter: I
Question
I tried solving this question, and I know the answer since it was given, but I could not get the same answer..
The answer has to be: pH: 1.50
First thin I tried was:
0.0116L * (4.7*10^-2 M OH) = 0.0005452 mol OH
0.0186L * (0.11M H) = 0.002046 mol H
At this point, I was not sure what I was suppose to do..
I tried using PH = pKa + log (base/acid), and also
PH = -log(H+)
but both did not give ph of 1.50.
I'll show you what I tried and please tell me what I did wrong ..
1) First trial
Ph = -log(H+)
(0.00206 mol H) - (0.0005452 mol OH) = 0.0015008 mol H left
(0.0015008mol H) / (0.0016L + 0.0186L) = 0.0497M H
PH = -log(0.0497) = 1.30
This is wrong since PH has to be 1.50 ..
2) Second Trial
OH + HA -> H20 + A
Since OH is limiting reactant, A becomes 0.0005452mol as well.
A(M) = (0.0005452 mol) / (0.0016L + 0.0186L) = 0.018M A
Ka = [H][A] / [HA]
(Here, I am not sure what I am suppose to use for [H] and [HA], do I use the given M or cauclated mol divided by the total volume?, also same question for in log(base/acid))
This eventually did not give me same answer as 1.50 ph..
Please help!
Explanation / Answer
11.6 mL of 4.7*10^-2 M Ca(OH)2
11.6 * 10^-3 * 4.7*10^-2 * 2 = 1.09 *10^-3 mols of OH
18.6 mL of 0.11 M HCl
18.6*10^-3*0.11 = 2.05*10^-3 mols of H+
2.05 - 1.09 = 0.96*10^-3 excess mol of H+
0.96*10^-3 /((11.6+18.6)*10^-3) = 3.18*10^-2 M
pH = -log(3.18*10^-2) = 3.45