1)
a)
\(CaC{l_2} + {K_2}C{O_3}\xrightarrow{{}}CaC{O_3} + 2KCl\)
Ion:
\(C{a^{2 + }} + C{O_3}^{2 - }\xrightarrow{{}}CaC{O_3}\)
b)
\(MgC{O_3} + 2HN{O_3}\xrightarrow{{}}Mg{(N{O_3})_2} + C{O_2} + {H_2}O\)
Ion:
\(MgC{O_3} + 2{H^ + }\xrightarrow{{}}M{g^{2 + }} + C{O_2} + {H_2}O\)
c)
\(FeS{O_4} + 2NaOH\xrightarrow{{}}Fe{(OH)_2} + N{a_2}S{O_4}\)
Ion:
\(F{e^{2 + }} + 2O{H^ - }\xrightarrow{{}}Fe{(OH)_2}\)
d)
\({(N{H_4})_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2N{H_3} + 2{H_2}O\)
Ion:
\(N{H_4}^ + + O{H^ - }\xrightarrow{{}}N{H_3} + {H_2}O\)
2)
Ta có:
\({n_{HCl}} = \frac{{1,825}}{{36,5}} = 0,05{\text{ mol}}\)
\( \to {C_{M{\text{ HCl}}}} = \frac{{0,05}}{{0,5}} = 0,1M = [{H^ + }] = [C{l^ - }]\)
\( \to pH = - \log [{H^ + }] = - \log (0,1) = 1\)
3)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,1.0,1 = 0,01{\text{ mol}}\)
\( \to {n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,02{\text{ mol;}}{{\text{n}}_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}} = 0,01{\text{ mol}}\)
\({n_{KOH}} = 0,1.0,22 = 0,022{\text{ mol = }}{{\text{n}}_{{K^ + }}} = {n_{O{H^ - }}}\)
Phản ứng xảy ra:
\({H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O\)
Vì \({n_{O{H^ - }}} > {n_{{H^ + }}} \to O{H^ - }\) dư.
Sau phản ứng dung dịch chứa:
\({n_{O{H^ - }{\text{ dư}}}} = 0,022 - 0,02 = 0,002{\text{ mol}}\)
\({n_{{K^ + }}} = 0,022{\text{ mol}}\)
\({n_{S{O_4}^{2 - }}} = 0,01{\text{ mol}}\)
Dung dịch sau phản ứng có thể tích:
\(V = 100 + 100 = 200ml = 0,2{\text{ lít}}\)
\( \to [O{H^ - }] = \frac{{0,002}}{{0,2}} = 0,01M\)
\([{K^ + }] = \frac{{0,022}}{{0,2}} = 0,11M\)
\([S{O_4}^{2 - }] = \frac{{0,01}}{{0,2}} = 0,05M\)
Ta có:
\(pOH = - \log [O{H^ - }] = - \log (0,01) = 2\)
\( \to pH=14 - pOH=14-2=12\)
