A 1.00 L solution contains 19.52 g of nitrous acid, HNO2. What mass of sodium nitrite, NaNO2, should be added to it to make a buffer with a pH of 2.56?

Question

Ka (HNO2) = 4.0 × 10–4.

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Answer 1 · 2018-05-01T06:04:51

Approx. $4*g$ of salt...

We use the buffer equation, for which...

$pH=pK_a+log_10{[[NO_2^(-)]]/[[HNO_2(aq)]]}$

$=underbrace(-log_10{4xx10^-4})_"3.40"+log_10{[[NO_2^(-)]]/[[HNO_2(aq)]]}$

$2.56=3.40+log_10{[[NO_2^(-)]]/[[HNO_2(aq)]]}$

CLEARLY $log_10{[[NO_2^(-)]]/[[HNO_2(aq)]]}=-0.84$

And so $[[NO_2^(-)]]/[[HNO_2(aq)]]=10^(-0.84)=0.145$ ..

...but $[HNO_2]=((19.52*g)/(47.01*g*mol^-1))/(1*L)=0.415*mol*L^-1$ ...

...therefore..
$[NO_2^-]=0.145xx0.415*mol*L^-1=0.0602*mol*L^-1$ .

And so...

$"mass of nitrous acid"=0.0602*mol*L^-1xx1*Lxx69.0*g*mol^-1=4.15*g$ ...

The acid is present in GREATER concentration in that we want to make the solution MORE acidic than the $pK_a$ of the acid.

Phew, arithmetic..