Question

A buffer consists of 0.120 m hno2 and 0.150 m nano2 at 25°c.

Answer 1

Answer:

A) pH of the buffer is 3.44

B) pH of the buffer solution is 3.37

Explanation:

Relation between K_{a} and pK_{a} is as follows.

$pK_{a} = -log (K_{a})$

. $pK_{a} = -log (K_{a}) = -log(4.5 \times 10^{-4}) = 3.347$

The relation between pH and  pK_{a} is as follows.

$pH = pK_{a} + log\frac{[conjugate base]}{[acid]}$

$= 3.347+ log \frac{0.15}{0.12} = 3.44$

pH of the buffer is 3.44.

b)

mol of HCl added = 11.6M *0.001 L = 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form$HNO_{2}$

No. of moles of

$NO^{-}_{2} = 0.15 M \times 1.0 L = 0.15 mol$

And, no. of moles of $HNO_{2} = 0.12 M \times 1.0 L$

                                              = 0.12 mol

after the reaction :  

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

                                         = (0.15 - 0.0116) mol

                                         = 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

                = (0.12 + 0.0116)

                = 0.1316 mol

As,

$K_{a} = 4.5 \times 10^{-4} pK_{a} = -log (K_{a}) = -log(4.5 \times 10^{-4}) = 3.347$

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

$pH = pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

           = 3.369

           ≅ 3.37

pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37