Finding out pH of the solution

[Dhamnekar Winod]

6.00 g acetic acid and 10.00 g sodium acetate are dissolved in enough water to make 1 L of solution.

a.What is the pH of the solution?
b.If 5 mL of 0.1 M HCl are added to the solution, what is the new pH?
c.If instead 5 mL of 1.0 M KOH are added to the original solution, what is the new pH?

I am working on these questions.

[Dhamnekar Winod]

Answer to question a.




\(pK_a\) for Ethanoic acid =4.77,\( pH= 4.77 + log(\frac{0.122 M }{0.1 M })=4.85\)

[ChenBeier]

The chemical reaction is wrong. Its only a mixture of acetic acid and sodium acetate. There is no reaction. You forget the lightning on the right side. Leftside you have a negative charge what gives thunder storm on the right side.
On the other hand the calculation of pH is correct because you consider the mixture.
Read about buffer solution in your text book or internet. This is relevant for b and c.

[Dhamnekar Winod]

Yes. Chemical reaction was wrongly written by me. Thanks for pointing out my mistake. I have corrected the chemical reaction now. is that correct?

[ChenBeier]

Yes I told you already. But how about b and c?

[Dhamnekar Winod]

Answer to question b.

\( pH = 4.77 + log (\frac{0.1215}{0.1005})=4.85\)

Is this answer correct?

[ChenBeier]

Yes correct

[Dhamnekar Winod]

Answer to question c.
Handerson-Hasselbalch equation = pKa + log ([conjugate base]/ [weak acid])

pH =pKa + log ([A⁻][HA]) ,

Hence, pH= 4.74 + log ( 0.127/0.095)= 4.84

Original Concentration of Vinegar - added concentration of 0.005M KOH =0.1 M -0.005 M KOH = 0.095 M

Original concentration of sodium acetate + added concentration of 0.005 KOH = 0.122 M + 0.005 M KOH = 0.127 M

[ChenBeier]

I dont know where you get the ml. Both acetic acid and sodium acetate are added as the mass. To get ml need specific gravity of 100% Acetic acid and sodium acetate is anyway a salt or need concentration before adding.