Buffer Solutions

The pH of the solution can be kept constant with the help of

Why constant pH solutions are needed?

Which of the following pairs will not constitute a buffer solution?

A buffer solution can be prepared by mixing equal volumes of

Among the following, the correct statement is

What is the concentration of sodium acetate that needs to be added to a $0.01 \mathrm{M}$ solution of acetic acid with equal volumes to give a solution of $\mathrm{pH}=5.5?$

$\left({\mathrm{pK}}_{\mathrm{a}}\text{ of }{\mathrm{CH}}_3\mathrm{COOH}=4.5\right)$

Choose the wrong statement from the following:

For preparing a buffer solution of $\mathrm{pH}=9$, by mixing ammonium chloride and ammonium hydroxide, the ratio of concentrations of salt and base should be_______ $\left({\mathrm{K}}_{\mathrm{b}}={10}^{-3}\right)$

A student prepares 2 L buffer solution of 0.3 M NaH₂PO₄ and 0.3 M Na₂HPO₄. The solution is divided in half between the two compartments (each containing 1 L buffer) of an electrolysis cell, using Pt electrodes. Assume that the only reaction is the electrolysis of water and the electrolysis is carried out for 200 min with a constant current of 0.965 A.

$\left[\text{Assume that}{\text{ pK}}_{\text{a}\left({\text{H}}_2{\text{PO}}_4^{\ominus }\right)}=\text{7.2}\right]$  [Given log 2.33 = 0.37]

pH at anode is

For next two question please follow the same

A student prepares 2 L buffer solution of 0.3 M NaH₂PO₄ and 0.3 M Na₂HPO₄. The solution is divided in half between the two compartments (each containing 1 L buffer) of an electrolysis cell, using Pt electrodes. Assume that the only reaction is the electrolysis of water and the electrolysis is carried out for 200 min with a constant current of 0.965 A.

$\left[\text{Assume that}{\text{ pK}}_{\text{a}\left({\text{H}}_2{\text{PO}}_4^{\ominus }\right)}=\text{7.2}\right]$  [Given log 2.33 = 0.37]

pH at cathode is

$100 \mathrm{mL}$ of $0.10 \mathrm{M}$ ${\mathrm{NH}}_4\mathrm{OH}$ mixed with $100 \mathrm{mL}$ of $0.05 \mathrm{M}$ $\mathrm{HCl}$ solution

The $\mathrm{pH}$ of $100 \mathrm{mL}$ of $0.10 \mathrm{M}$ $\mathrm{NaOH}$ mixed with $100 \mathrm{mL}$ $0.10 \mathrm{M}$ ${\mathrm{CH}}_3\mathrm{COOH}$ solution is

The number of moles of an acid or base added to one litre of the buffer solution so as to change its $\mathrm{pH}$ by one unit is called_____of the buffer.

Blood is a buffer of H2CO3 and HCO3- with pH of 7.4. What will be the ratio of HCO3- to H2CO3 in blood.

The ionisation constant of $\mathrm{NH}_{4}^{+}$ in water is $5.6 \times 10^{-10} \mathrm{at}$ $25^{\circ} \mathrm{C} .$ The rate constant for the reaction of $\mathrm{NH}_{4}^{+}$ and $\mathrm{OH}^{-}$ to form $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ at $25^{\circ} \mathrm{C}$ is $3.4 \times 10^{10}$ litre

$\mathrm{mol}^{-1} \mathrm{sec}^{-1} .$ If equilibrium constant of water at $25^{\circ} \mathrm{C}$ is $1.8 \times 10^{-16},$ thenThe $\mathrm{pH}$ of a mixture of

$0.1 \mathrm{M} {\mathrm{NH}}_4\mathrm{OH}$ and
$0.1\mathrm{M} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ solution is $\left[{\mathrm{K}}_{\mathrm{a}} \text{of }{\mathrm{NH}}_4^{+}\text{ is} 5.6\times {10}^{-10} \text{and} \log 1.78=0.25\right]$ 

A buffer of acetic acid $({\mathrm{pK}}_{\mathrm{a}}=4.8)$ with sodium acetate, when ${\mathrm{CH}}_3\mathrm{COOH}$ and ${\mathrm{CH}}_3\mathrm{COONa}$ are present in equivalent amounts, has a$\mathrm{pH}$ range of

The buffer capacity is equal to:

Consider the titration of weak monobasic acid HA using 0.1 M NaOH. If initially 100 mL of 0.1 M HA is taken then find the volume of NaOH (in mL) added so that pH of mixture become equal to 5.



 

The number of grams/weight of $\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}$ required to be added to 3 litres of 0.01 M $\mathrm{N}{\mathrm{H}}_3$ to prepare the buffer of $\mathrm{p}\mathrm{H}=9.45$ at temperature 298 K is $(K_b\mathrm{ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{ }{\mathrm{N}\mathrm{H}}_3 \mathrm{i}\mathrm{s}\mathrm{ }1.85\times {10}^{-5})$

In the neutralization process of H₃PO₄ and NaOH, the number of buffers formed will be-