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Question: Calculate the amount of ammonium chloride required to dissolved in \(500\;mL\) water to have \(pH = ...

Calculate the amount of ammonium chloride required to dissolved in 500  mL500\;mL water to have pH=4.5(Kb  for  NH4OH=1.8×105)pH = 4.5\left( {{K_b}\;{\rm{for}}\;N{H_4}OH = 1.8 \times {{10}^{ - 5}}} \right).

Explanation

Solution

As we know that, ammonium chloride is a compound of ammonia and ammonium hydrogen chloride. We can determine the pH value by using the formula.
pH=12(pKWlogCpKb)pH = \dfrac{1}{2}\left( {p{K_W} - \log C - p{K_b}} \right)

Complete step by step solution
According to the question ammonium chloride requires dissolving in 500  ml500\;ml and its pH value is given.
As we know that NH4ClN{H_4}Cl is a salt of weak base and strong acid. Using the formula, we can get the valve of ammonium chloride to dissolve. This formula represents the relation between the pH and pKbp{K_b}.Here pKwp{K_w} is the water autoionization, and pKbp{K_b} is the base dissociation. On substituting these values, we can find the pH value.
pH=12(pKWlogCpKb) 4.5=12[14logC+log(1.8×105)] 9=14logC4.74 logC=0.26........................................(1) C=x26.25,  where,  Mass  of  ammonium  chloride2=52.52=26.25 pH = \dfrac{1}{2}\left( {p{K_W} - \log C - p{K_b}} \right)\\\ \Rightarrow 4.5 = \dfrac{1}{2}\left[ {14 - \log C + \log \left( {1.8 \times {{10}^{ - 5}}} \right)} \right]\\\ \Rightarrow9 = 14 - \log C - 4.74\\\ \Rightarrow \log C = 0.26........................................(1)\\\ \Rightarrow C = \dfrac{x}{{26.25}},\;{\rm{where,}}\;\dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{ammonium}}\;{\rm{chloride}}}}{{\rm{2}}} = \dfrac{{52.5}}{2} = 26.25
Where x is the amount of NH4ClN{H_4}Cl dissolved in 500  ml500\;ml of water. When we take both side log we get value of log C, putting the value of log C in equation 2 we get,
log  c=log(x26.25)..........................................(2)\log \;c = \log \left( {\dfrac{x}{{26.25}}} \right)..........................................(2)
Here we find the value of log c, on solving these we can find the value of x. Which is the amount of NH4ClN{H_4}Cl dissolved in 500  ml500\;ml of water.
log(x26.25)=0.26 logxlog26.25=0.26 logx=0.26+1.4191=1.6791 x=Antilog(1.6791) x=47.76 \Rightarrow \log \left( {\dfrac{x}{{26.25}}} \right) = 0.26\\\ \Rightarrow \log x - \log 26.25 = 0.26\\\ \log x = 0.26 + 1.4191 = 1.6791\\\ \Rightarrow x = Anti\log \left( {1.6791} \right)\\\ \Rightarrow x = 47.76
Therefore, the amount of NH4ClN{H_4}Cl dissolved is 47.76  g47.76\;g.
NH4ClN{H_4}Cl Solution is acidic in nature because it reacts with water and produces ammonium hydroxide and hydrochloric acid. Ammonium hydroxide is a weak base. So it is partially ionized in aqueous solution and HClHCl is a strong acid, As a result, in the solution the number of H+{H^ + } ions are more than OHO{H^ - } ions.
We know that the acidic and basic property of substances is measured in terms of pH. It is a measurement of the hydrogen ion concentration. And we know that it is a negative log of hydrogen ion concentration.

Note:
We know that when ammonium chloride is dissolved in water it decomposes in its ion like NH4+  and  ClN{H_4 }^+ \;and\;C{l^ - }. It can back in crystallization form.