Chemistry, 13.07.2021 04:00 erikloza12pdidtx
Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).
Answers: 1
is 4.49.
= 0.10 M
![Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}](/tpl/images/1393/0831/fa913.png)
is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.![K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}](/tpl/images/1393/0831/ba8e8.png)
![[H_{3}O^{+}] = 3.2 \times 10^{-5}](/tpl/images/1393/0831/f5dd0.png)
![pH = -log [H^{+}]](/tpl/images/1393/0831/8d00e.png)
![pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49](/tpl/images/1393/0831/392b2.png)
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Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)...
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