Answer to Question #146967 in General Chemistry for liam donohue

Question

An aqueous solution started at pH= 4.58 but with the addition of an acid reduced to 1.96. At the end of the total V= 124.55 ml. Solid Li2CO3 was added (MM= 73.89G/mol) to reestablish the original pH.
What mass of bass is required

Answer 1

Mass of $Li_2CO_3=x$

Molar mass of $Li_2CO_3=73.89g/mol$

Volume of solution$=0.1245L$

$K_w=1.0\times 10^{-4}$

pH of solution$=1.96$

pH=$-log [H^+][H^+]=10^{-pH}$

$10^{-4}=1.0\times 10^{-4}$

Dissociation equation is;

$Li_2CO_3\leftrightharpoons Li^{2+}+CO^{2-}_3$

$1.0\times 10^{-4}x=$ $K_a$ $=[Li^{2+}][CO_3^{2-}]\over Li_2CO_3$

$4.6\times 10^{-4}(1.96)\over 1.0\times 10^{-4}$ $=9.016M$

$[CO_3^{2-}]=9.016M$

Mass of $Li_2CO_3=?$

$Mass=$ $Mol\over L$ $]CO_3^{2-}$

$=$ $9.016mol\over L$ $\times 0.1245L$

$=1.1225CO_3^{2-}$

$mLi_2CO_3=1.1225 mol\times 73.89g/mol=82.94g$

$\therefore$ Mass of $Li_2CO_3=82.94g$