Question

1 L of $0.1{\text{ M}}$ NaOH, 1 L of $0.1{\text{ M}}$ KOH, and 2 L of 0.05{\text{ M}}$$$Ba{(OH)_2}$ are mixed together. The final concentration of the solution is: A.0.01{\text{ M}}$B.$0.01{\text{ N}}$C.$0.1{\text{ N}}$D.$0.001{\text{ M}}$$

Answer 1

A mixture of two or more substances where one or more solutes are dissolved uniformly in a solvent is known as solution. A solution is a homogeneous mixture. We shall multiply the concentration with the volume of each compound and add them to get the final concentration.

Complete step by step answer:
In chemistry, concentration is the number or amount of a constituent present in the mixture or solution divided by the total volume of mixture or solution. The different types of concentration are as follows,
Mass concentration: The mass concentration is defined as the mass of the particle or substance present in the mixture divided by the volume of the mixture. The SI unit of mass concentration is $\dfrac{{{\text{kg}}}}{{{{\text{m}}^{\text{3}}}}}$.
Molar concentration: The molar concentration is defined as the number of moles of the substance present in the mixture divided by the volume of mixture. The SI unit of molar concentration is $\dfrac{{{\text{mol}}}}{{{{\text{m}}^{\text{3}}}}}$.
Number concentration: The number concentration is defined as the number of constituents present in the mixture divided by the total volume of the mixture. The SI unit of number concentration is ${{\text{m}}^{{\text{ - 3}}}}$.
Volume concentration: The volume concentration is defined as the volume of a substance in a mixture divided by the volume of total mixture. This quantity is dimensionless.
Let us assume volume and normality of NaOH as ${V_1}$ and ${N_1}$, that of KOH as ${V_2}$ and ${N_2}$, and that of $Ba{(OH)_2}$ as ${V_3}$ and ${N_3}$.
Total volume(V) = ${V_1}$ + ${V_2}$ + ${V_3}$
${\text{V}} = 1 + 1 + 2 = 4{\text{ L}}$
n factor of NaOH = 1
n factor of KOH = 1
n factor of $Ba{(OH)_2}$ = 2
Normality, N is equal to the product of n factor and molarity.
Hence, ${N_1} = 1 \times 0.1 = 0.1$
${N_2} = 1 \times 0.1 = 0.1$
${N_3} = 2 \times 0.05 = 0.1$
Now, NV = ${N_1}{V_1}$ + ${N_2}{V_2}$ + ${N_3}{V_3}$
Substituting the values,
$N \times 4 = 0.1 \times 1 + 0.1 \times 1 + 0.1 \times 2$
$N = \dfrac{{0.4}}{4} = 0.1{\text{ N}}$
Hence the final concentration of the solution is$0.1{\text{ N}}$.

Therefore, the correct answer is option C.
Note:
NaOH is used in making soap, detergent etc.
KOH is used in liquid fertilizers, potassium soaps etc.
$Ba{(OH)_2}$ is used in pesticides and manufacture of glass.