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Question: Calculate pH of a solution whose \[100\;ml\]contains \[0.2g\;NaOH\] dissolved in it? A. \[10.699\]...

Calculate pH of a solution whose 100  ml100\;mlcontains 0.2g  NaOH0.2g\;NaOH dissolved in it?
A. 10.69910.699
B. 11.69911.699
C. 12.69912.699
D. 13.69913.699

Explanation

Solution

The pH is equal to negative logarithm to the base ten of x, where, x is the hydrogen ion concentration unit is in moles per litre/s. When solute is base which is dissolved in neutral solvent here it is water, then the value of pH should be greater than seven as the base has a value greater than seven. The question here is what is the exact value of pH, for that the equation of pH is used.

Complete step by step solution
The range of pH scale is in between 00 to 1414 including 00and 1414.The pH value is divided in three parts first, from 00 to 77 indicates the acidic nature of solution. Second, from 77 to 1414 indicates the basic nature of the solution.

The intermediate value that is 77 indicates neutral nature of solution such as pure water or neutral substance which will not give hydrogen ion nor hydroxyl ion. The pH value of Ammonia or Sodium Hydroxide is in the range 77-1414, which reflects their basic nature.

Whereas acidic substances like vinegar CH3COOHC{H_3}COOH, stomach acid (HClHCl), battery acid, etc, show pH range from 00-77 is shown by acidic substances like vinegar CH3COOHC{H_3}COOH, stomach acid (HClHCl), battery acid, etc, as per its acidic nature.

The concentration is expressed as follows:
Concentration(Molarity)=moles  of  soluteVolume  of  solution(in  litres)Concentration{\rm{ }}\left( {Molarity} \right) = \dfrac{{moles{\rm{\; }}of{\rm{\; }}solute}}{{Volume{\rm{\; }}of{\rm{\; }}solution{\rm{ }}\left( {in{\rm{ \;}}litres} \right)}}

Moles  of  solute=  mass  of  sodium  hydroxidemolecular      mass =0.240 =0.005  molMoles{\rm{ \;}}of{\rm{\; }}solute = {\rm{ \;}}\dfrac{{mass\;of\;sodium \;hydroxide}}{{molecular\;\;\;mass}}\\\ = \dfrac{{0.2}}{{40}}{\rm{ }}\\\ = {\rm{ }}0.005\;mol

Volume of solution=100ml=0.1litres = 100ml = \,0.1\,litres.

[OH]=nvolume =0.0050.1 =0.05M\left[ {OH - } \right] = \dfrac{n}{{volume}}\\\ = \dfrac{{0.005}}{{0.1}}\\\ = 0.05M pOH=log10  [OH]=log10(0.05)   pOH=1.301 pH+pOH=14 Hence,pH=12.699{pOH = - lo{g_{10}}\;\left[ {OH - } \right] = - lo{g_{10}}\left( {0.05} \right)\;}\\\ {pOH{\rm{ }} = {\rm{ }}1.301}\\\ {pH{\rm{ }} + {\rm{ }}pOH{\rm{ }} = {\rm{ }}14}\\\ {Hence,{\rm{ }}pH{\rm{ }} = {\rm{ }}12.699}

**Therefore, the correct answer is option (C) 12.69912.699.

Note: **
In a few experiments the pH meter is used which indicates the exact value of pH same as a thermometer is used to measure the value of temperature, and a galvanometer is used for current.