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
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
Use the Henderson-Hasselbalch equation.
pH = pKa + log(base)/(acid)
base = sodium acetate.
acid = acetic acid.
Substitute and solve.
Nope. I don't know how you came up with 5.45.
pH = 4.74 + log(0.15/0.2) = 4.6 or so.
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
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.
No I'm sorry I just used this as the example of the H H equation
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.
This what I did log(1.76 x 10-5) +log(0.150/0.200)
= 4.879
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