Question

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What is the approximate pH of a 0.05 M solution of HSCN that has a Ka of 10-4?
HSCN -> H+ + SCN-

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The equation for calculating pH is pH = -log [H+]
Now we need to find H+ and solve.

[HSCN] = 0.05 M the molarity
Ka = [H+][SCN-]/[HSCN] is the Ka equation

HSCN → H+ + SCN-
0.05 M 0M 0M
-x +x +x

[HSCN] = (0.05 – x)M
[H+][SCN-] = x^2

Now we solve for H+
Ka = [x^2]/[0.05-x]
(10^-4)^2/(0.05-10^-4) inputting the Ka for x
= √(10^-4)^2/√(0.0499)
= (10^-4)/0.223383
= 0.000448
or 4.48*10^-4

pH = -log 4.48*10^-4
pH = 3.348722

The approximate pH of a 0.05 M solution of HSCN that has a Ka of 10-4 is 3.35

Answer 1

Are you familiar with the simplification concept when calculating hydronium ion concentrations or weak acids in pure water? Basically, it says that in the I.C.E. Table, at equilibrium, if the concentration of weak acid divided by the Ka value is > 100, then one can drop the 'x' in the wk acid concentration term and not have sinificant error in the concentration of hydronium ion. That is,

HSCN <=> H^+ + SCN^-
0.05 ---------- 0 --------- 0
-x ---------- +x --------- +x
0.05 - x ----- x --------- x
~0.05M
'x' may be dropped as Conc/Ka > 100

[H^+] = Sqr-Root(Ka[acid])
----- = Sqr-root[(1x10^4)(0.05)]M
----- = 2.24 x 10^-4M

pH = -log[H^+] = -log(2.24 x 10^-4)
-- = 2.65

Answer 2

Thank you!

Answer 3

You are most welcome. Doc (aka Olereb48)