Calculation of pH and hydroniumn ion
Chemistry

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The pH of an aqueous solution of 0.324 M hydrocyanic acid is
The hydronium ion concentration of an aqueous solution of 0.442 M hydrofluoric acid (K_{a} = 7.20E4) is
1. hydrocyanic acid (HCN) has a ka value of 6.2E10
The equilibirum rxn is given by...
HCN + H2O <> H3O+ + CN
I.....0.324 M............................0 M............0M
C... x....................................+x...............+x
E.0.324 M  x..........................x..................x
The equilibrium expression becomes...
6.2E10 = x^2 / (0.324 M  x)
Using either quadratic formula or method of succesive approximations yields
x = 1.4172904E5 M
Since the hydronium ion concentration is given by x and since pH = log [H3O+], the pH is given by...
pH = log (1.4172904E5)
pH = 4.85 (with sig figs in mind)
2.
The equilibirum rxn is given by...
HF + H2O <> H3O+ + F
I.....0.442 M............................0 M............0M
C... x....................................+x...............+x
E.0.442 M  x..........................x..................x
The equilibrium expression becomes...
7.20E4= x^2 / (0.442 M  x)
Using either quadratic formula or method of succesive approximations yields
x = 1.7482912 M
Since the hydronium ion concentration is given by x the hydronium ion concentration is simply...
[H3O+] = 1.75 E2 M or 0.0175 M (with sigfigs in mind)
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