Question

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?

Ka (HNO2) = 4.0 × 10–4.

Answer 1

Approx. `4*g` of salt...

Explanation:

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..