# A 0.495 M solution of nitrous acid, HNO2, has a pH of 1.83.Calculate the pH of the solution formed by adding 1.0 g of NaNO2 (molar mass = 69.0 g·mol–1) to750 mL of 0.0125 M solution of nitrous...

A 0.495 M solution of nitrous acid, HNO2, has a pH of 1.83.Calculate the pH of the solution formed by adding 1.0 g of NaNO2 (molar mass = 69.0 g·mol–1) to750 mL of 0.0125 M solution of nitrous acid.

*print*Print*list*Cite

Nitrous acid is a weak acid dissociating in water solution as:

`HNO_2 stackrel (larr)(->) H^+ + NO_2^-`

Let a be the initial concentration of `HNO_2` , and x, the amount dissociated,

Then `[HNO_2 ] ` = (a-x),` [H^+ ]` =x and `[NO_2^- ]` =x

Dissociation constant of `HNO_2` , i.e. K_a = `[HNO_2 ]/ ([H^+ ][NO_2^- ])`

=`x^2/(a-x)`

A 0.495M solution of `HNO_2` has a pH of 1.83.

`rArr [H+] = 10^-1.83` = 0.014791 M

Putting the values in the expression for `K_a` , and remembering `[NO_2^-] = [H^+]` , we get

`K_a = (0.014791)^2/(0.495-0.014791)` = `0.000456 ` = `4.56 × 10^-4`

`rArr pKa=-log(4.56 × 10^-4)` = `3.341`

In the second solution, an aqueous solution of a weak acid and its salt (with a strong base) is made. This is an acidic buffer solution, whose pH can be calculated from the Henderson-Hasselbalch equation,

`pH = pKa + log([A^-]/[HA])`

Putting the values from second condtion,

`pH = 3.341+log([1000/(69*750)]/[0.0125])` =3.341+0.1892 = 3.53

Hence, the pH of the solution formed by adding 1.0 g of NaNO2 to750 mL of 0.0125 M solution of nitrous acid is **3.53**.