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Question: \(5.6{\text{ grams}}\) of \({\text{KOH}}\) (M.wt \( = 56\)) is present in \(1{\text{ litre}}\) of so...

5.6 grams5.6{\text{ grams}} of KOH{\text{KOH}} (M.wt =56 = 56) is present in 1 litre1{\text{ litre}} of solution. Its pH is:
A) 1
B) 13
C) 14
D) 0

Explanation

Solution

The negative logarithm of the H+{{\text{H}}^ + } ion concentration in the solution is known as the pH of the solution. Initially calculate the molarity of the solution then calculate the pH.

Formula Used: Number of moles (mol)=Mass (g)Molar mass (g mol−1){\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}
Molarity (M)=Number of moles of solute (mol)Volume of solvent (L){\text{Molarity (M)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Volume of solvent (L)}}}}
pOH=−log⁡[OH−]{\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]
pH+pOH=14{\text{pH}} + {\text{pOH}} = {\text{14}}

Complete step by step answer:
Calculate the number of moles of KOH{\text{KOH}} in 5.6 grams5.6{\text{ grams}} of KOH{\text{KOH}} using the equation as follows:
Number of moles (mol)=Mass (g)Molar mass (g mol−1){\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}
Substitute 5.6 grams5.6{\text{ grams}} for the mass of KOH{\text{KOH}}, 56 g mol−156{\text{ g mo}}{{\text{l}}^{ - 1}} for the molar mass of KOH{\text{KOH}} and solve for the number of moles of KOH{\text{KOH}}. Thus,
Number of moles of KOH=5.6 g56 g mol−1{\text{Number of moles of KOH}} = \dfrac{{{\text{5}}{\text{.6 g}}}}{{56{\text{ g mo}}{{\text{l}}^{ - 1}}}}
Number of moles of KOH=0.1 mol{\text{Number of moles of KOH}} = 0.1{\text{ mol}}
Thus, the number of moles of KOH{\text{KOH}} in 5.6 grams5.6{\text{ grams}} of KOH{\text{KOH}} are 0.1 mol0.1{\text{ mol}}.
Calculate the molarity of the solution using the equation as follows:
Molarity (M)=Number of moles of solute (mol)Volume of solvent (L){\text{Molarity (M)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Volume of solvent (L)}}}}
Substitute 0.1 mol0.1{\text{ mol}} for the number of moles of KOH{\text{KOH}}, 1 L1{\text{ L}} for the volume of the solvent and solve for the molarity of KOH{\text{KOH}}. Thus,
Molarity of KOH=0.1 mol1 L{\text{Molarity of KOH}} = \dfrac{{0.1{\text{ mol}}}}{{{\text{1 L}}}}
Molarity of KOH=0.1 M{\text{Molarity of KOH}} = 0.1{\text{ M}}
Thus, the molarity of the solution is 0.1 M0.1{\text{ M}}
Calculate the pOH of the solution as follows:
We know that KOH{\text{KOH}} is a strong base. Thus, KOH{\text{KOH}} dissociates completely. Thus,
[K+]=[OH−]=0.1 mol\left[ {{{\text{K}}^ + }} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right] = 0.1{\text{ mol}}
We know that the negative logarithm of the hydroxide ion concentration is known as pOH. Thus,
pOH=−log⁡[OH−]{\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]
Substitute 0.1 M0.1{\text{ M}} for the concentration of hydroxide ion and solve for the pOH. Thus,
pOH=−log⁡[0.1 M]{\text{pOH}} = - \log \left[ {0.1{\text{ M}}} \right]
pOH=1{\text{pOH}} = 1
Thus, the pOH of the solution is 1.
Calculate the pH of the solution using the equation as follows:
pH+pOH=14{\text{pH}} + {\text{pOH}} = {\text{14}}
Rearrange the equation for the pH of the solution. Thus,
pH=14−pOH{\text{pH}} = {\text{14}} - {\text{pOH}}
Substitute 1 for the pOH and solve for the pH. Thus,
pH=14−1{\text{pH}} = {\text{14}} - 1
pH=13{\text{pH}} = {\text{13}}
Thus, the pH of the solution is 13.

Thus, the correct option is (B).

Note: The pH of the solution is 13. This indicates that the pH is greater than 7. If the pH is greater than 7 the solution is basic in nature. If the pH is less than 7 the solution is acidic in nature. If the pH is equal to 7 the solution is neutral.