Science
Calculation of pH and hydroniumn ion

### Question Description

The pH of an aqueous solution of 0.324 hydrocyanic acid is

The hydronium ion concentration of an aqueous solution of 0.442 M hydrofluoric acid (Ka = 7.20E-4) is

1. hydrocyanic acid (HCN) has a ka value of 6.2E-10

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...

6.2E-10 = x^2 / (0.324 M - x)

Using either quadratic formula or method of succesive approximations yields

x = 1.4172904E-5 M

Since the hydronium ion concentration is given by x and since pH = -log [H3O+], the pH is given by...

pH =  -log (1.4172904E-5)

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...

7.20E-4= x^2 / (0.442 M - x)

Using either quadratic formula or method of succesive approximations yields

x = 1.748291-2 M

Since the hydronium ion concentration is given by x the hydronium ion concentration is simply...

[H3O+] = 1.75 E-2 M  or 0.0175 M (with sigfigs in mind)

Steve_R944 (190)
Cornell University
Review

Anonymous
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Anonymous
Thanks as always for the good work!

Anonymous
Excellent job

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