InterviewSolution
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InterviewSolution
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Calculate the concentration of all species present in `0.010M H_2SO_4` solution. `(K_(a_2)=1.3xx10^(-2))` |
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Answer» Strategy: Sulphuric acid `(H_2SO_4)` is an example of a diprotic acid because each unit of the acid yields two `H^(+)` ions, in two separate steps: `H_2SO_4(aq.)rarrH^(+)(aq.)+HSO_4^(-)(aq.)` `HSO_4^(-)(aq.)hArr H^(+)(aq.)+SO_4^(2-)(aq.)` `H_2SO_4` is a strong electrolyte (or strong acid) as the first step of ionization is complete, but `HSO_4^(-)` is weak acid, we need a double arrow to represent its incomplete ionization. Because the first ionization step of `H_2SO_4` is complete, we read the concentration for the first step from the balanced equation. The second ionization step is not complete. Thus, we write the ionization equation, the `K_(a_2)` expression, and the algebratic representations of equilibrium concentrations. Then we substitute into `K_(a_2)` for `H_2SO_4`. Solution: Summarize the changes in the first stage of ionization which is complete: `{:(,H_(2)SO_(4)(aq.)overset(100%)rarrH^(+)(aq.)+HSO_(4)^(-)(aq.)),("Initial (M)"," 0.010 0.00 0.00"),("Change (M)"," -0.010 +0.010 +0.010"),("Final (M)", bar(" 0.00 0.010 0.010 ")):}` For the second stage of ionization (which is not complete), we proceed as for a weak monoprotic acid. Step 1: Let `x=C_(HSO_4^(-))` that ionizes (or the concentration in `mol L^(-1)` of `H^(+)` and `SO_4^(2-)` produced by the ionization of `HSO_4^(-)`). The total concentration of `H^(+)` ions at equilibrium must be the sum of the `H^(+)` ion concentration produced in the first and second steps. So we represent the equilibrium concentrations as `{:(,HSO_(4)^(-)(aq.)hArrH^(+)(aq.)+SO_(4)^(2-)(aq.)),("Initial (M)"," 0.10 0.010 0.00"),("Change (M)"," -x +x +x"),("Equilibrium (M)", bar(" (0.010-x) (0.010+x)" " x ")):}` Step 2: Substitution into the ionization constant expression for `K_(a_2)` gives `K_(a_2)=(C_(H^+)C_(SO_4^(2-)))/(C_(HSO_4^(-))` `1.3xx10^(-2)=((0.010+x)(x))/((0.010-x))` Since the `K_(a_2)` of `H_2SO_4` is quite large, x cannot be neglected. We must solve the quadratic equation, which simplies to `x^2+0.023x-1.3xx10^(-4)=0` Applying the quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)` we have `x=(-0.023+-sqrt((0.023)^2-4(1)(-1.3xx10^(-4))))/(2(1))` `=4.7xx10^(-3)M` `:. (C_(H^+))_(2nd)=C_(SO_4^(2-))=4.7xx10^(-3)M` The concentrations of different species in `0.010M H_2SO_4` are `C_(H_2SO_4)~~0M`, `C_(HSO_4^(-))=(0.010-x)` `=(0.010-4.7xx10^(-3))` `=5.3xx10^(-3)` `C_(H^(+))=(0.010+x)` `=(0.010+4.7xx10^(-3))` `=0.015` `C_(SO_4^(2-))=4.7xx10^(-3)` `C_(OH^(-))=(K_w)/(C_(H^+))=(1.0xx10^(-14))/(0.015)=6.7xx10^(-13)` |
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