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Finding out pH of the solution
Posted: Wed Apr 07, 2021 7:09 am
by 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.
Re: Finding out pH of the solution
Posted: Thu Apr 08, 2021 10:14 pm
by Dhamnekar Winod
Answer to question a.
\(pK_a\) for Ethanoic acid =4.77,\( pH= 4.77 + log(\frac{0.122 M }{0.1 M })=4.85\)
Re: Finding out pH of the solution
Posted: Thu Apr 08, 2021 11:39 pm
by 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.
Re: Finding out pH of the solution
Posted: Fri Apr 09, 2021 12:54 am
by 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?
Re: Finding out pH of the solution
Posted: Fri Apr 09, 2021 3:54 am
by ChenBeier
Yes I told you already. But how about b and c?
Re: Finding out pH of the solution
Posted: Fri Apr 09, 2021 9:38 am
by Dhamnekar Winod
Answer to question b.
\( pH = 4.77 + log (\frac{0.1215}{0.1005})=4.85\)
Is this answer correct?
Re: Finding out pH of the solution
Posted: Fri Apr 09, 2021 10:20 am
by ChenBeier
Yes correct
Re: Finding out pH of the solution
Posted: Fri Apr 09, 2021 11:22 pm
by 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
Re: Finding out pH of the solution
Posted: Sat Apr 10, 2021 2:06 am
by 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.