Answer to Question #146967 in General Chemistry for liam donohue

Question #146967
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
1
Expert's answer
2020-12-06T10:26:55-0500

Mass of Li2CO3=xLi_2CO_3=x

Molar mass of Li2CO3=73.89g/molLi_2CO_3=73.89g/mol

Volume of solution=0.1245L=0.1245L

Kw=1.0×104K_w=1.0\times 10^{-4}

pH of solution=1.96=1.96

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

104=1.0×10410^{-4}=1.0\times 10^{-4}

Dissociation equation is;

Li2CO3Li2++CO32Li_2CO_3\leftrightharpoons Li^{2+}+CO^{2-}_3

1.0×104x=1.0\times 10^{-4}x= KaK_a =[Li2+][CO32]Li2CO3=[Li^{2+}][CO_3^{2-}]\over Li_2CO_3

4.6×104(1.96)1.0×1044.6\times 10^{-4}(1.96)\over 1.0\times 10^{-4} =9.016M=9.016M

[CO32]=9.016M[CO_3^{2-}]=9.016M

Mass of Li2CO3=?Li_2CO_3=?

Mass=Mass= MolLMol\over L ]CO32]CO_3^{2-}

== 9.016molL9.016mol\over L ×0.1245L\times 0.1245L

=1.1225CO32=1.1225CO_3^{2-}

mLi2CO3=1.1225mol×73.89g/mol=82.94gmLi_2CO_3=1.1225 mol\times 73.89g/mol=82.94g

\therefore Mass of Li2CO3=82.94gLi_2CO_3=82.94g


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