Question Number 154330 by DELETED last updated on 17/Sep/21
Answered by DELETED last updated on 17/Sep/21
![Jawab: Dik=K_1 =8,318×10^(−7) K_2 =5,248×10^(−10) [H^+ ]=10^(−8) ΣCO_2 =2,3 mMol [CO_3 ^(−2) ]=((DIC)/([1+(([H^+ ])/k_2 )+(([H^+ ]^2 )/(k_1 .k_2 ))])) =((2,3)/(1+((10^(−8) )/(10^(−9,28) ))+(([10^(−8) ]^2 )/(10^(−6,8) ×10^(−9,28) )))) =((2,3)/(1+10^(1,28) +10^(0,8) )) =((2,3)/(1+19,054+6,309)) =((2,3)/(26,36))=0,0873//](https://www.tinkutara.com/question/Q154331.png)
$$\mathrm{Jawab}: \\ $$$$\mathrm{Dik}=\mathrm{K}_{\mathrm{1}} =\mathrm{8},\mathrm{318}×\mathrm{10}^{−\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}_{\mathrm{2}} =\mathrm{5},\mathrm{248}×\mathrm{10}^{−\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{H}^{+} \right]=\mathrm{10}^{−\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Sigma\mathrm{CO}_{\mathrm{2}} =\mathrm{2},\mathrm{3}\:\mathrm{mMol} \\ $$$$\left[\mathrm{CO}_{\mathrm{3}} ^{−\mathrm{2}} \right]=\frac{\mathrm{DIC}}{\left[\mathrm{1}+\frac{\left[\mathrm{H}^{+} \right]}{\mathrm{k}_{\mathrm{2}} }+\frac{\left[\mathrm{H}^{+} \right]^{\mathrm{2}} }{\mathrm{k}_{\mathrm{1}} .\mathrm{k}_{\mathrm{2}} }\right]} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2},\mathrm{3}}{\mathrm{1}+\frac{\mathrm{10}^{−\mathrm{8}} }{\mathrm{10}^{−\mathrm{9},\mathrm{28}} }+\frac{\left[\mathrm{10}^{−\mathrm{8}} \right]^{\mathrm{2}} }{\mathrm{10}^{−\mathrm{6},\mathrm{8}} ×\mathrm{10}^{−\mathrm{9},\mathrm{28}} }} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2},\mathrm{3}}{\mathrm{1}+\mathrm{10}^{\mathrm{1},\mathrm{28}} +\mathrm{10}^{\mathrm{0},\mathrm{8}} } \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2},\mathrm{3}}{\mathrm{1}+\mathrm{19},\mathrm{054}+\mathrm{6},\mathrm{309}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2},\mathrm{3}}{\mathrm{26},\mathrm{36}}=\mathrm{0},\mathrm{0873}//\: \\ $$