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21
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by AWK on Yesterday at 09:32:45 AM »
You have an excess of NaOH. What does it mean?
22
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by OkPluto on Yesterday at 09:27:45 AM »
I did the first one:
[HCN] = [NaOH] = [NaCN]= 0.033 M
pH = 1/2(14 + 8.96 + log_10(0.033)) = 10.74

But the second one is still wrong
[HCN] = 0.025 M
[NaOH] = 0.0375 M
[HCN] = 0.025 M
pH = 1/2(14 + 8.96 + log_10(0.025)) = 10.68
24
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by OkPluto on Yesterday at 08:55:34 AM »
Which is that equation? That was the only i saw during lessons
25
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by AWK on Yesterday at 08:37:33 AM »
Hydrolysis of salt uses other equation. Calculate salt concentration.
Correct stoichiometry will show you the correct way of calculation
26
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by OkPluto on Yesterday at 08:34:31 AM »
Ok, but these comments are not helpful...
There is something that i can't do, underline only where am i wrong will not make me progress in this problem..
27
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by AWK on Yesterday at 08:26:58 AM »
1 - hydrolysis of the salt
2 - wrong stoichiometry.
28
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by OkPluto on Yesterday at 08:06:52 AM »
the acid is HCN and its conjugate base is NaCN. I already understood that but i don't know how to solve the problem yet
29
Undergraduate General Chemistry Forum / Re: Determine pH
« Last post by Borek on Yesterday at 07:40:50 AM »
You are making the same mistake in both problems. Henderson-Hasselbalch equation uses concentrations of acid and its conjugate base. What are acid and its conjugate base in this case?
30
Undergraduate General Chemistry Forum / Determine pH
« Last post by OkPluto on Yesterday at 06:37:39 AM »
I'm having trouble with this 2 problems:

Find the pH of the solution obtained by mixing 20 mL of NaOH 0.05 M and 10 mL of HCN 0.1 M
(pKa HCN = 8.96)
Solution is 10.74
I found that there are 1*10^-3 mol of both NaOH and HCN in solution and consequently there are 1*10^-3 mol of NaCN
so i calculate pH = pKa + log_10(1/1) = pkA = 8.96 that is a different from 10.74


Find the pH of the solution obtained by mixing 30 mL of 0.05 M NaOH and 10 mL of 0.1 M HCN
(pKaHCN = 8.96)
Solution is 12.10
I found that there are 1*10^-3 mol NaOH and 1.5*10^-3 mol of HCN in solution and consequently there are 5*10^-4 mol of NaCN
so i calculate pH = pKa + log_10(1.5/0.5) = pkA + log_10(3) = 9.43 that is different from 12.10
Can anyone help me understand where am i wrong?
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