Buffer Solutions

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 $\mathrm{pH}=5.5?$

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

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 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.

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.



 

$pH$ of an acid buffer is given by

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-

Which one of the following mixture does not act as a buffer solution-

100 mL of a solution contains 2g of acetic acid and 3g of sodium acetate providing $K_a=1.8\times {10}^{-5}$ , then choose the correct option.

In a basic buffer, $0.0025 \mathrm{mole}$ of ${\mathrm{NH}}_4\mathrm{Cl}$ and $0.15 \mathrm{mole}$ of ${\mathrm{NH}}_4\mathrm{OH}$ are present. The $\mathrm{pH}$ of the solution will be $\left({\mathrm{pK}}_{\mathrm{a}}\right)=4.74.$ $[\log 10=1\&\log 6=0.78]$

A solution of $0.1 \mathrm{N}$ $\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}$ and $0.1 \mathrm{N}$ $\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}$ has $\mathrm{pH} 9.25$. Then, $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ of $\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}$ is

Solution of 0.1 N $\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}$ and 0.1 N $\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}$ has pH 9.25. Then, $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ of $\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}$ is

In a buffer solution having $0.02 \mathrm{M}\mathrm{ }{\mathrm{NH}}_4\mathrm{OH}$ and $0.05 \mathrm{M}\mathrm{ }{\mathrm{NH}}_4\mathrm{Cl}$, $0.01 \mathrm{M}\mathrm{ }\mathrm{NaOH}$ is added. The change in $\mathrm{pH}$ value of buffer will be $({\mathrm{pK}}_{{\mathrm{NH}}_4\mathrm{OH}}=4.74)$

A solution is prepared by mixing equal volumes of $0.4\mathrm{ }\mathrm{M}\mathrm{ }{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}$ and $0.2\mathrm{ }\mathrm{M}\mathrm{ }{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$. ${\mathrm{K}}_{\mathrm{a}}$ for ${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}={10}^{–6}$. The $\mathrm{p}\mathrm{H}$ of the resulting solution would be -

The $\mathrm{p}\mathrm{H}$ of a solution obtained on mixing $50\mathrm{ }\mathrm{m}\mathrm{L}$ of $0.1\mathrm{ }\mathrm{M}$ formic acid with $150\mathrm{ }\mathrm{m}\mathrm{L}$ of $0.02\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ is:

$\left({\mathrm{K}}_{\mathrm{a}}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=1.8\mathrm{ }\times \mathrm{ }{10}^{–4};\mathrm{ }\log 1.8=0.255\right)$