Propanoic acid (HC3H8O2) has a Ka of 1.3*10^-5. if you are titrating a 25.00 mL sample of 0.355 M propanoic acid with 0.525 M NaOH, calculate the pH before the titration has started.
HC3H8O2 dissociates as:
HC3H8O2 -----> H+ + C3H8O2-
0.355 0 0
0.355-x x x
Ka = [H+][C3H8O2-]/[HC3H8O2]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.3*10^-5)*0.355) = 2.148*10^-3
since c is much greater than x, our assumption is correct
so, x = 2.148*10^-3 M
So, [H+] = x = 2.148*10^-3 M
use:
pH = -log [H+]
= -log (2.148*10^-3)
= 2.6679
Answer: 2.67
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