Question

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

Answer 1

Hint: The negative logarithm of the ${{\text{H}}^ + }$ 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: ${\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}$
${\text{Molarity (M)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Volume of solvent (L)}}}}$
${\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]$
${\text{pH}} + {\text{pOH}} = {\text{14}}$

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