A sample of Na_2CO_3 cdot H_2O weighing 0.62 | vedclass

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(C) $1$. Calculate the moles of $Na_2CO_3 \cdot H_2O$: The molar mass is $124 \ g/mol$. Moles $= 0.62 \ g / 124 \ g/mol = 0.005 \ mol$. $2$. Calculate

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(C) $1$. Calculate the moles of $Na_2CO_3 \cdot H_2O$: The molar mass is $124 \ g/mol$. Moles $= 0.62 \ g / 124 \ g/mol = 0.005 \ mol$. $2$. Calculate the equivalents of $(NH_4)_2SO_4$: Equivalents $= N 	imes V(L) = 0.1 \ N 	imes 0.1 \ L = 0.01 \ eq$. $3$. The reaction is: $Na_2CO_3 + (NH_4)_2SO_4 ightarrow (NH_4)_2CO_3 + Na_2SO_4$. $4$. $1 \ mol$ of $Na_2CO_3$ reacts with $1 \ mol$ of $(NH_4)_2SO_4$. Since $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,it provides $2$ equivalents per mole. Thus,$0.005 \ mol$ of $Na_2CO_3$ provides $0.01 \ eq$. $5$. Since $0.01 \ eq$ of $Na_2CO_3$ reacts with $0.01 \ eq$ of $(NH_4)_2SO_4$,the resulting solution contains $(NH_4)_2CO_3$,which is a salt of a weak acid and a weak base. The pH of such a salt is given by $pH = 7 + 0.5(pK_a - pK_b)$. For $(NH_4)_2CO_3$,$pK_a$ of $HCO_3^-$ is $\approx 10.3$ and $pK_b$ of $NH_4OH$ is $\approx 4.75$,resulting in a slightly basic solution.

$A$ sample of $Na_2CO_3 \cdot H_2O$ weighing $0.62 \ g$ is added to $100 \ mL$ of $0.1 \ N \ (NH_4)_2SO_4$ solution. What will be the nature of the resulting solution?

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