Henderson Hassleback Equation

Q. A solution is prepared by mixing $0.01$ mole each of ${\text{H}}_2{\text{CO}}_3,{\text{NaHCO}}_3,{\text{Na}}_2{\text{CO}}_3$, and $\text{NaOH}$ in $100\text{mL}$ of water. $\text{pH}$ of the resulting solution is

[Given : ${\text{pKa}}_1$ and ${\text{pKa}}_2$ of ${\text{H}}_2{\text{CO}}_3$ are $6.37$ and 10.32, respectively; $\log 2=0.30$]

Q.

To what volume one litre of 0.1M acetic acid solution diluted so that the pH of the solution becomes double of the initial ? ( $K_a=1.8\times {10}^{-5}$)

1. 5.5 x 10⁵L

2. 3.6 x 10⁸ L

3. 8 x 10⁴ L

4. 2.4 x 10³ L

Q.

Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. ($K_a$ of acetic acid = $1.8\times {10}^{-5}$)

1. 5.6

2. 4.74

3. 12.8

4. 8.62

Q.

The pH of an aqueous solution at ${25}^{\circ }C$ having 0.1 M NaOH and 0.3 M to acetic acid ($p{K}_a$ = 4.76) would be nearly

1. 4.25

2. 5.05

3. 4.45

4. 4.75

Q.

They yield of acid amide in the reaction, $RCOCl+N{H}_3\to RCON{H}_2$ , is maximum when

1. Acid chloride and ammonia are treated in equimolar ratio

2. Acid chloride and ammonia are treated in 2 : 1 molar ratio

3. All the three gives nearly similar result

4. Acid chloride and ammonia are treated in 1 : 2 molar ratio

Q.

To 1L solution containing 0.1 mol each of $N{H}_3$ and $N{H}_{4}Cl,0.05$ mol of NaOH is added. The change in pH will be $(p{K}_{b}forN{H}_3=4.74)$

1. -0.48

2. 0.48

3. 0.30

4. -0.30

Q.

pH of 0.5 M $Ba(CN{)}_2$ solution $(p{K}_{b}ofC{N}^{-}=9.30)$ is :

1. 8.35

2. 3.35

3. 9.35

4. 9.50

Q.

A weak base BOH is titrated with a strong acid HA. When 10ml of HA is added, the pH is found to be 9.00 and when 25 ml is added, pH is 8.00. The volume of the acid required to reach the equivalence point

1. 50 ml

2. 35 ml

3. 40 ml

4. 30 ml

Q.

The base imidazole has $K_b$ of $1.0\times {10}^{-7}$. What volumes of 0.02 M HCl and 0.02 M imidazole should be mixed to make 150 mL of a buffer of pH 7 ?

1. 75 ml and 75 ml

2. 50 mL and 100 mL

3. 100 ml and 50 ml

4. 60 ml and 90 ml

Q.

$0.1mol$ of methyl amine ($K_b$ = $5\times {10}^{-4}$) is mixed with $0.08mol$ of $HCl$ and the solution is diluted to $1litre$. The hydrogen ion concentration of the resulting solution is:

1. $\left[H^{+}\right]$ = $3.68\times {10}^{-11}M$

2. $\left[H^{+}\right]$ = $1.79\times {10}^{-11}M$

3. $\left[H^{+}\right]$ = $6.53\times {10}^{-11}M$

4. $\left[H^{+}\right]$ = $8\times {10}^{-11}M$