the other half of the buffer + 6.0mL 1 M NaOH Calculate the expected pH of solut

Question

the other half of the buffer + 6.0mL 1 M NaOH Calculate the expected pH of solutions prepared as given below. Enter the results into the Calculated pH blocks on the second page of the report sheet. Buffer containing 4 mL 6 M acetic acid (pK-4.74) +3.3 g sodium acetate trihydrate (MM= 136 g/mol) +water to make up 50 ml. rtl: pra tag (2e24? rn :?46? Half (25 mL) of the buffer+ 6.0 mL 1 MHCL. 4.71 The other half of the buffer +6.0 mL 1 M NaOH. 24 16 To the buffer with HCl add another 10 mL HCl for a total of 16.0 mL 1 MHCI. To the buffer with NaOH add another 10 mL NaOH for a total of 16.0 mL 1 M NaOH

Explanation / Answer

Question: the other half of the buffer + 6.0mL 1 M NaOH

4 mL of 6 M acetic acid
Moles of acetic acid = 0.004 L x 6 M = 0.024 moles

3.3 g of sodium acetate trihydrate
Moles of sodium acetate trihydrate = 3.3 g / (136 g/mol) = 0.024 moles

Total volume was 50 mL then half of it was taken, so the moles will also be half
So, moles are
Moles of acetic acid = 0.024 moles / 2 = 0.012 moles
Moles of sodium acetate trihydrate = 0.024 moles / 2 = 0.012 moles

6.0 mL of 1 M NaOH
So, moles of NaOH = 0.006 L x 1 M = 0.006 moles

When 0.006 moles of NaOH is added, it will react with 0.006 moles of acetic acid to form 0.006 moles of sodium acetate. So, moles of acetic acid will decrease and moles of sodium acetate will increase.

Moles after adding NaOH are
Moles of acetic acid = 0.012 moles – 0.006 moles = 0.006 moles
Moles of sodium acetate trihydrate = 0.012 moles + 0.006 moles = 0.018 moles

Total volume = 25 mL = 0.025 L

So, the concentration terms are
[acetic acid] = 0.006 moles / 0.025 L = 0.24 M
[sodium acetate trihydrate] = 0.018 moles / 0.025 L = 0.72 M

pKa of acetic acid = 4.74

Using Henderson-Hesselbalach equation;

pH = pKa + log { [salt] / [acid] }

pH = pKa + log { [sodium acetate] /[acetic acid] }
      = 4.74 + log (0.72 / 0.24)
      = 4.74 + log (3)
      = 4.74 + 0.48
      = 5.22

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