Question

A solution of formic acid has a pH of 2.70, calculate the initial concentration of formic acid?

`K_a=1.8xx10^-4` for formic acid

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My work:

Answer 1

`underbrace([HC(=O)OH]=0.0241*mol*L^-1)_"INITIALLY"`

Explanation:

We address the equilibrium...

`HC(=O)OH(aq)+H_2O(l)rightleftharpoonsHCO_2^(-) + H_3O^+`

And at equilibrium...`K_a=([HCO_2^(-)][H_3O^+])/([HC(=O)OH(aq)])=1.80xx10^-4`

But `pH=-log_10[H_3O^+]=2.70`...

And so `[H_3O^+]=[HCO_2^-]=10^(-2.70)*mol*L^-1=`

And so AT EQUILIBRIUM...`1.80xx10^-4=({10^(-2.70)}^2)/([HC(=O)OH(aq)])`

`[HC(=O)OH(aq)]_"at equilibrium"=({10^(-2.70)}^2)/(1.80xx10^-4)`

`-=0.0221*mol*L^-1`...but this is the equilibrium value....and `[H_3O^+]=1.995xx10^-3*mol*L^-1` by definition...

And since the hydronium ion is PRESUMED to derive from the formic acid...`[HC(=O)OH]_"initially"=(0.0221+1.995 xx 10^(-3))*mol*L^-1=0.0241*mol*L^-1`...the degree of dissociation was miniscule....

Pleas check my 'rithmetik...