At ${25^0}C$ ,the value of $p{K_b}({K_b}$ being the dissociation constant at a base ) for $N{H_3}$ in aqueous solution is $4.7$ . What is the $pH$ of $0.1M$ aqueous solution of $N{H_4}Cl$ with $0.01M$ $N{H_3}$ (approximately)?
A. $8.3$
B. $9$
C. $9.5$
D. $10$
Answer
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Hint: Ammonia acts as a weak base in aqueous solution. The pH of standard ammonia is about 11. Ammonia is base which reacts in water and gives positively charged ammonium ion and negatively charged hydroxide ion ( $N{H_3} + {H_2}O \rightleftharpoons N{H^ + }_4 + O{H^ - }$ ).
Complete answer:
The base dissociation constant ${K_b}$ is given as, ${K_b} = \dfrac{{[N{H^ + }].[O{H^ - }]}}{{[N{H_3}]}}$.We also know some important formulas that is $pH = 14 + \log ([O{H^ - }])$ , ${pH} + {pOH} = 14$ (by this formula we subtract ${pH}$ value from 14 and get ${pOH}$ value ) and ${pOH} = p{K_b} + \log \dfrac{{[salt]}}{{[weak base]}}$. pH is the measure of the concentration of hydrogen ion in an aqueous solution and ${pOH}$ measures the hydroxide ion concentration of the solution . The base dissociation constant measures the dissociation of base into its components ions in aqueous solution . Here in the problem the value of dissociation constant of base is given that is $p{K_b} = 4.7$
Now the concentration of the salt $N{H_4}Cl$ is $0.1M$ . (given)
Again the concentration of the weak base ($N{H_3}$) is given that is $0.001M$.
$ \Rightarrow {pOH} = 4.7 + \log \dfrac{{[0.1]}}{{[0.001]}}$
$ \Rightarrow {pOH} = 5.7$
$ \Rightarrow {pH} = 14 - 5.7$
$ \Rightarrow {pH} = 8.3$
So the correct option is A that is 8.3. The $pH$ of $0.1M$ aqueous solution of $N{H_4}Cl$ with $0.01M$ $N{H_3}$ is 8.3.
Note: Ammonia can be manufactured and it is present in nature naturally. At equilibrium concentration the ratio between concentration of ammonium cations and hydroxide anions to the ammonia is equal to the base dissociation constant. With the help of some important formulas which we have learnt so far we had found out the $pH$ of $0.1M$ aqueous solution of $N{H_4}Cl$ with $0.01M$ $N{H_3}$. A large value of $p{K_b}$ shows that there is a large dissociation of a base. Thus low $p{K_b}$ means that the given base is a strong base and vice-versa.
Complete answer:
The base dissociation constant ${K_b}$ is given as, ${K_b} = \dfrac{{[N{H^ + }].[O{H^ - }]}}{{[N{H_3}]}}$.We also know some important formulas that is $pH = 14 + \log ([O{H^ - }])$ , ${pH} + {pOH} = 14$ (by this formula we subtract ${pH}$ value from 14 and get ${pOH}$ value ) and ${pOH} = p{K_b} + \log \dfrac{{[salt]}}{{[weak base]}}$. pH is the measure of the concentration of hydrogen ion in an aqueous solution and ${pOH}$ measures the hydroxide ion concentration of the solution . The base dissociation constant measures the dissociation of base into its components ions in aqueous solution . Here in the problem the value of dissociation constant of base is given that is $p{K_b} = 4.7$
Now the concentration of the salt $N{H_4}Cl$ is $0.1M$ . (given)
Again the concentration of the weak base ($N{H_3}$) is given that is $0.001M$.
$ \Rightarrow {pOH} = 4.7 + \log \dfrac{{[0.1]}}{{[0.001]}}$
$ \Rightarrow {pOH} = 5.7$
$ \Rightarrow {pH} = 14 - 5.7$
$ \Rightarrow {pH} = 8.3$
So the correct option is A that is 8.3. The $pH$ of $0.1M$ aqueous solution of $N{H_4}Cl$ with $0.01M$ $N{H_3}$ is 8.3.
Note: Ammonia can be manufactured and it is present in nature naturally. At equilibrium concentration the ratio between concentration of ammonium cations and hydroxide anions to the ammonia is equal to the base dissociation constant. With the help of some important formulas which we have learnt so far we had found out the $pH$ of $0.1M$ aqueous solution of $N{H_4}Cl$ with $0.01M$ $N{H_3}$. A large value of $p{K_b}$ shows that there is a large dissociation of a base. Thus low $p{K_b}$ means that the given base is a strong base and vice-versa.
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