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.
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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^- ])`
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.
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