pH question ?
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH.
A solution is made by mixing 6.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?
- davidLv 72 months ago
NaOH is a strong base so I will use it. Any other strong base would do the same.
Ka = 5.61×10−6
6.00 mmol HA + 2.00 mmol NaOH .. obviously the Acid is in excess .... 4.00 mmol HA remain unreacted
2.00mmol NaA are formed ... The problem is that you gave NO VOLUMES ..... the formula for Ka uses molar concentrations, not just moles. === U will use v for the unknown volume.
HA + NaOH --> NaA + H2O
NaA --> Na+ + A-
HA <--> H+ + A-
. . . . . init . . . . . change . . . . . . equikub
[HA] . . 4.00/v . . - x/v . . . . . . 4/v - x/v
[H+] . . 0 . . . . . . +x/v . . . . . . . . x/v
[A-] . . 2.00/v . . . + x/v . . . . . . . 2/v + x/v
= = = = = = = = = = = = = = = = = = = = = = = =
Ka = [H+][A-] / [HA]
5.61x10^-6 = (x/v)((4-x)/v) / ((2+x)/v)
5.61x10^-6 = (x/v)((4-x)) / ((2+x)) <<< 2 unknowns x and v ... but only one equation .... impossible to solve.
... assuming v = 1 ... then I'll continue
5.61x10^-6 = (x)((4-x)) / (2+x) <<<<<< assume x is 'small' and round
5.61x10^-6 = (x)((4)) / (2) = 2x
x = 2.805x10^-6 = [H+]
pH = -log 2.805x10^-6 = 5.55 <<< round as appropriate