# Question #c5de8

May 7, 2015

All but $N a I \text{/} H I$.

A buffer must contain one of two conjugate pairs in about equal ammounts

• A weak acid and its conjugate base, or
• A weak base and its conjugate acid.

So, right off the bat, if you spot a strong acid or a strong base in one of the pairs, it will automatically mean that you're not dealing with a buffer.

Upon inspection of the options given to you, you can see that you are indeed dealing with a strong acid in one of the pairs; more specifically, the last pair contains hydroiodic acid, $H I$, so that solution cannot act as a buffer.

Now for the other three.

• $K C N \text{/} H C N$

This pair contains hydrogen cyanide, $H C N$, which is a weak acid. Moreover, the solution also contains potassium cyanide, $K C N$, a salt of the cyanide ion, which is the conjugate base of hydrogen cyanide.

${\underbrace{H C {N}_{\left(a q\right)}}}_{\textcolor{b l u e}{\text{weak acid")) + H_2O_((a)) rightleftharpoons H_3O_((aq))^(+) + underbrace(CN_((aq))^(-))_(color(green)("conjugate base}}}$

Since it contains both a weak acid, and its conjugate base, presumably in equal amounts, this solution will act as a buffer.

• $N {a}_{2} H S {O}_{4} \text{/} N a {H}_{2} S {O}_{4}$

This one is a little more difficult to spot. You're dealing with two salts, monosodium phosphate, $N a {H}_{2} P {O}_{4}$, and disodium phosphate, $N {a}_{2} H P {O}_{4}$.

However, in aqueous solution, the two salts will dissociate completely to give

$N a {H}_{2} P {O}_{4 \left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + {H}_{2} P {O}_{4 \left(a q\right)}^{-}$

and

$N {a}_{2} H P {O}_{4 \left(a q\right)} \to 2 N {a}_{\left(a q\right)}^{+} + H P {O}_{4 \left(a q\right)}^{2 -}$

Notice that you've discovered dihydrogen phosphate, ${H}_{2} P {O}_{4}^{-}$, which is a weak acid, and hydrogen phosphate, $H P {O}_{4}^{2 -}$, which is its conjugate base.

As a result, this solution will also act as a buffer.

• $N {H}_{3} \text{/} N {H}_{4} N {O}_{3}$

This time you're dealing with ammonia, $N {H}_{3}$, which is a weak base, and ammonium nitrate, which is the salt of ammonia's conjugate acid, the ammonium ion, $N {H}_{4}^{+}$.

${\underbrace{N {H}_{3 \left(a q\right)}}}_{\textcolor{b l u e}{\text{weak base")) + H_2O_((l)) rightleftharpoons OH_ ((aq))^(-) + underbrace(NH_(4(aq))^(+))_(color(green)("conjugate acid}}}$

This solution will also act as a buffer.

Buffers are all about acids and bases, so you need to know how to recognize them, which ones are strong, which ones are weak, and what conjugate acids and conjugate bases are.