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Salt solutions Quiz 2

chemistry

 The formula of pH for acid and basic buffer respectively are:

 The formulas can be derived by seeing the dissociation and it is also called the Henderson Hasselbalch equation.

  $ \text{The pH of an acidic buffer is }pH=p{{K}_{a}}+\log \dfrac{[Salt]}{[Acid]}, \text{and the pH of a basic buffer is }\\ pH=p{{K}_{b}}+\log \dfrac{[Base]}{[Salt]}$.

  If $C{{H}_{3}}COOH$ and $C{{H}_{3}}COONa$ are mixed as buffers, then the common ion effect will be seen in which ion?

  For the common ion effect, the solution contains an excess of the common ion and less amount of ${{H}_{3}}{{O}^{+}}$ ions.

 When the compounds are mixed, buffer action takes place. Let us see the equations: $\\ C{{H}_{3}}COOH+{{H}_{2}}O\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}_{3}}{{O}^{+}} \\ C{{H}_{3}}COONa\to C{{H}_{3}}CO{{O}^{-}}+N{{a}^{+}} \\ $ So, we can see that the common ion effect is seen in $C{{H}_{3}}CO{{O}^{-}}$.

  A salt of strong acid and the weak base has a dissociation constant ${{K}_{b}}=1.8\times {{10}^{-5}}$, among the options, find the correct pH of the solution if 0.1 M concentration is taken.

 The formula for finding the pH is $pH=7-\dfrac{1}{2}[p{{K}_{b}}+\log c]$

 $ \text{We know that:}\\  pH=7-\dfrac{1}{2}[p{{K}_{b}}+\log c] \\  \Rightarrow pH=7-\dfrac{1}{2}[-\log (1.8\times {{10}^{-5}})+\log 0.10] \\  \Rightarrow pH=7-\dfrac{1}{2}(5-0.26-1)=5.12 $

  Find the concentration of $C{{a}^{2+}}$ ions in aqueous solution, if it is given that ${{k}_{sp}}(M{{X}_{2}})=4\times {{10}^{-12}}$, where $CaC{{l}_{2}}$ is the salt.

 Find a relation between solubility and solubility product to find the answer.

$CaC{{l}_{2}}\rightleftharpoons C{{a}^{2+}}+2C{{l}^{-}}, \text{we take concentration of }\\ C{{a}^{2+}}=S;2C{{l}^{-}}=2S, \text{so the }{{K}_{sp}}=S{{(2S)}^{2}}=4{{S}^{3}}.\text{So,}\\ S=\sqrt[3]{\dfrac{{{K}_{sp}}}{4}}=\sqrt[3]{\dfrac{4\times {{10}^{-2}}}{4}}=1\times {{10}^{-4}}M$

  A salt of a weak base and strong acid has ${{K}_{b}}=1.8\times {{10}^{-5}}$, and it has pH as 4.5, when it is mixed in ${{H}_{2}}O$. Find the mass of the salt that is needed, given that there is the availability of 500mL of water. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/14c6fd20-fb47-48a2-8091-0c87cfddef111278917615933641624.png"/>

 From the given data, find ${{K}_{h}},h$ and the concentration. Then find the mass in grams.

$ \text{The concentration of }{{H}^{+}}={{10}^{-pH}}={{10}^{-4.5}}=3.162\times {{10}^{-5}}M.\\ \text{If }C \text{ is concentration and } h \text{ is dissociation constant, then }\\ {{K}_{h}}=C{{h}^{2}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}=\dfrac{{{10}^{-14}}}{1.8\times {{10}^{-5}}}=5.5\times {{10}^{-10}}, \\ \text{and we can see that h} =\dfrac{{{K}_{h}}}{Ch}=\dfrac{{{K}_{h}}}{[{{H}^{+}}]}=\dfrac{5.5\times {{10}^{-10}}}{3.162\times {{10}^{-5}}}=1.74\times {{10}^{-5}}.\\ \text{As, }[{{H}^{+}}]=C\times h, \text{we have }C=\dfrac{[{{H}^{+}}]}{h}=\dfrac{3.162\times {{10}^{-5}}}{1.74\times {{10}^{-5}}}=1.8mol\,{{L}^{-1}}.\\ \text{So, 500mL of water shall contain half the amount,} \text{ i.e 0.9 mole, and the mass in grams is }\\ 0.9\times 53.5=48.15g$.