Given
[HNO2] = 0.406 M
To calculate pH of a weak acid we have to use ICE table
HNO2 | H2O <===> | H3O+ | NO2- | |
I | 0.406 M | - | 0 | 0 |
C | -x | - | +x | +x |
E | 0.406-x | - | x | x |
Ka of HNO2 = 4.5 x 10-4
but
Neglecting denominator x (0.406-x = 0.406)
x = 0.0135 M
[H3O+] = x = 0.0135 M
[NO2-] = x= 0.0135 M
[HNO2] = 0406 - x = 0.406- 0.0135 = 0.392 M
Formula
pH = -log[H3O+]
pH = -log (0.0135)
pH = 1.87
------------------------------------------
Given,
[HC9H7O4] = 0.0170 M
Using ICE table to solve for pH
HC9H7O4 | H2O <==> | C9H7O4- | H3O+ | |
I | 0.0170 M | - | 0 | 0 |
C | -x | - | +x | +x |
E | 0.0170 -x | - | x | x |
Given
here we are asked to solve by quadratic equation so we cannot neglect x
it is in the form of ax2 +bx +c =0
formula to solve for x
Concentration wont be in negative
so X= 0.00224 M
[H3O+] = 0.00224 M
[C9H7O4-] = x = 0.00224 M
[HC9H7O4] = 0.0170- x = 0.0170- 0.00224 = 0.0148 M
pH = -log[H3O+]
pH = -log(0.00224)
pH = 2.65
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