# A 100*mL volume of 2.0*mol*L^-1 hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

Feb 16, 2017

The equivalent weight of the metal is $100.0 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

First we need a chemical equation:

$M C {O}_{3} \left(s\right) + 2 H C l \left(a q\right) \rightarrow M C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

$\text{Moles of HCl} = 0.100 \cdot L \times 2 \cdot m o l \cdot {L}^{-} 1 = 0.200 \cdot m o l$

Given the equation, the $\text{metal carbonate}$ represents $0.100 \cdot m o l$, and thus, given the equation, the equivalent weight of the $\text{metal carbonate}$ is $100.0 \cdot g \cdot m o {l}^{-} 1$. This equivalent weight is CONSISTENT with a formula of $C a C {O}_{3}$.

Oct 21, 2017

Equivalent weight of metal is 20 g/eq

#### Explanation:

For HCl
Molarity = Normality

Gram equivalents of HCl = Normality × Volume in litres
= 2 N × 100 × 10^-3 L
$= 0.2$ eq

For complete neutralisation
Gram equivalents of HCl = Gram equivalents of metal carbonate

Equivalent weight of metal carbonate $= \frac{\textrm{W e i g h t o f m \eta l c a r b o n a t e}}{\textrm{G r a m \equiv a \le n t s o f m \eta l c a r b o n a t e}}$

$= \frac{10 g}{0.2 \textrm{e q}} = 50$ g/eq

Equivalent weight of carbonate (CO_3^(2-)) = (12 + (3 × 16))/2 = 30 g/eq

Equivalent weight of metal carbonate = Equivalent weight of metal + Equivalent weight of carbonate

50 g/eq = x + 30 g/eq
x = 20 g/eq

∴ Equivalent weight of metal is 20 g/eq

Oct 21, 2017

The number of moles of $H C L$ reacted with Carbonate salt is

$= \text{molarity of acid" xx " volume of acid in L}$

$= 2 m o l \text{/} L \times 0.1 L = 0.2 m o l$

So the mass of $H C l$ reacted $= 0.2 \times 36.5 g$, where molar mass of $H C l = 36.5 g \text{/} m o l$

Now we know from the law of equivalent proportion that the ratio of masses of two reacting substances for neutralization is equal to the ratio of their equivalent masses.

Hence we can say

"Equivalent mass of metal carboate" /("Equivalent mas of "HCl)="Reacting mass of metal carboate" /("Equivalent mas of "HCl)

$\implies \left(\text{Equivalent mass of metal (x=?)+ equivalent mass of "CO_3^(2-))/("Equivalent mas of } H C l\right) = \frac{0.2 \times 36.5}{36.5}$

$\implies \left(x + 30 \text{g/equiv")/(36.5"g/equiv}\right) = \frac{10 g}{0.2 \times 36.5 g}$

$\implies \textcolor{red}{x = 20 \text{g/equiv}}$

Equivalent mass of CO_3^(2-)= (" formula mass of " CO_3^(2-))/"valency "
$= \frac{12 + 3 \times 16}{2} = 30 \text{g/equiv}$