Calculation of pH and hydroniumn ion

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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

Mar 9th, 2015

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)



Mar 9th, 2015

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