calculate the pH of a buffer solution containing 0.200M acetic acid, HC2H3O2, plus 0.150M sodium acetate. The dissociation constant for HC2H3O2 is 1.76 x 10-5

Question

I don't think our instructor gave enough notes on these things Im kind of winging it from the book. Could you please show steps

Answer 1

Use the Henderson-Hasselbalch equation.
pH = pKa + log(base)/(acid)
base = sodium acetate.
acid = acetic acid.
Substitute and solve.

Answer 2

Ok DrBob I think I got it after all the plugging in my final answer is 5.45 I hope

Answer 3

Nope. I don't know how you came up with 5.45.
pH = 4.74 + log(0.15/0.2) = 4.6 or so.

Answer 4

pH = pKa + log ([A-]/[HA])

pH = pKa + log ([C2H3O2-] / [HC2H3O2])

pH = -log (1.8 x 10-5) + log (0.50 M / 0.20 M)

pH = -log (1.8 x 10-5) + log (2.5)

pH = 4.7 + 0.40

pH = 5.1
I used this example
And I still got it wrong it should have been 4.75 + 0.12 = 4.87 or do I use just 4.7 + 0.12

Answer 5

Where are you getting the mL? There aren't any in the post. The original post says base = 0.15 and acid = 0.2 and those are the numbers that you substitute.

Answer 6

No I'm sorry I just used this as the example of the H H equation

Answer 7

Disregard that. You don't have mL listed; however, you should have the log term base/acid as 0.15/0.20 and not 0.50/0.2.

Answer 8

This what I did log(1.76 x 10-5) +log(0.150/0.200)
= 4.879

Answer 9

No. I THINK you are dividing 0.2/0.15 to get 1.33 and log 1.33 = about 0.124 and that + 4.74 = about 4.4.86.

Follow this.
pH = 4.74 + log (0.15/0.20)
pH = 4.74 + log(0.75)
pH = 4.74 + (-.125)
pH = 4.74 - 0.125 = 4.61

Answer 10

DrBob I want to thank you SOOOOO much for all the help you give. I had never heard of the H H equation. And I also appreciate you patience

Answer 11

Good luck.

Answer 12

Keq= (.200)(1.76 x 10^-5) / .150
work the math then with that answer, you do the Ph log and will get your answer