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##### Undergraduate General Chemistry Forum / Re: Determine pH

« Last post by**AWK**on

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**Yesterday**at 09:32:45 AMYou have an excess of NaOH. What does it mean?

June 01, 2020, 02:17:17 AM

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You have an excess of NaOH. What does it mean?

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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

[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

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Which is that equation? That was the only i saw during lessons

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Hydrolysis of salt uses other equation. Calculate salt concentration.

Correct stoichiometry will show you the correct way of calculation

Correct stoichiometry will show you the correct way of calculation

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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..

There is something that i can't do, underline only where am i wrong will not make me progress in this problem..

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1 - hydrolysis of the salt

2 - wrong stoichiometry.

2 - wrong stoichiometry.

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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

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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?

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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?

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?