Question Number 154321 by DELETED last updated on 17/Sep/21
Answered by DELETED last updated on 17/Sep/21
![•pK_1 =6,08 →−log K_1 =−log 10^(−6,08) →K_1 =10^(−6,08) •pK_2 =9,28 →−log K_2 =−log 10^(−9,28) •pH=8→[H^+ ]=10^(−8) salinitas 35 ? T=10° C ? P=1 atm ? yg merah bikin pak ngah bingung bel seumur hidup baru dpt soal spt ini bel.ada contoh teorinya gak?](Q154323.png)
$$\bullet\mathrm{pK}_{\mathrm{1}} =\mathrm{6},\mathrm{08}\:\rightarrow−\mathrm{log}\:\mathrm{K}_{\mathrm{1}} =−\mathrm{log}\:\mathrm{10}^{−\mathrm{6},\mathrm{08}} \\ $$$$\rightarrow\mathrm{K}_{\mathrm{1}} =\mathrm{10}^{−\mathrm{6},\mathrm{08}} \\ $$$$\bullet\mathrm{pK}_{\mathrm{2}} =\mathrm{9},\mathrm{28}\:\rightarrow−\mathrm{log}\:\mathrm{K}_{\mathrm{2}} =−\mathrm{log}\:\mathrm{10}^{−\mathrm{9},\mathrm{28}} \\ $$$$\bullet\mathrm{pH}=\mathrm{8}\rightarrow\left[\mathrm{H}^{+} \right]=\mathrm{10}^{−\mathrm{8}} \\ $$$$\mathrm{salinitas}\:\mathrm{35}\:? \\ $$$$\mathrm{T}=\mathrm{10}°\:\mathrm{C}\:? \\ $$$$\mathrm{P}=\mathrm{1}\:\mathrm{atm}\:? \\ $$$$\mathrm{yg}\:\mathrm{merah}\:\mathrm{bikin}\:\mathrm{pak}\:\mathrm{ngah}\:\mathrm{bingung}\:\mathrm{bel} \\ $$$$\mathrm{seumur}\:\mathrm{hidup}\:\mathrm{baru}\:\mathrm{dpt}\:\mathrm{soal} \\ $$$$\mathrm{spt}\:\mathrm{ini}\:\mathrm{bel}.\mathrm{ada}\:\mathrm{contoh}\:\mathrm{teorinya} \\ $$$$\mathrm{gak}? \\ $$$$ \\ $$$$ \\ $$
Answered by DELETED last updated on 17/Sep/21
![•CO_2 +H_2 O⇋HCO_3 ^− +H^+ K_1 =(([HCO_3 ^− ][H^+ ])/([CO_2 ][H_2 O])) →K_1 =10^(−6,8) •HCO_3 ^− +H^+ ⇋2H^+ +CO_3 ^(−2) K_2 =(([2H^+ ]^2 [CO_3 ^(−2) ])/([HCO_3 ^− ][H^+ ])) →K_2 =10^(−9,28)](Q154325.png)
$$\bullet\mathrm{CO}_{\mathrm{2}} +\mathrm{H}_{\mathrm{2}} \mathrm{O}\leftrightharpoons\mathrm{HCO}_{\mathrm{3}} ^{−} +\mathrm{H}^{+} \\ $$$$\:\:\mathrm{K}_{\mathrm{1}} =\frac{\left[\mathrm{HCO}_{\mathrm{3}} ^{−} \right]\left[\mathrm{H}^{+} \right]}{\left[\mathrm{CO}_{\mathrm{2}} \right]\left[\mathrm{H}_{\mathrm{2}} \mathrm{O}\right]}\:\rightarrow\mathrm{K}_{\mathrm{1}} =\mathrm{10}^{−\mathrm{6},\mathrm{8}} \\ $$$$\bullet\mathrm{HCO}_{\mathrm{3}} ^{−} +\mathrm{H}^{+} \leftrightharpoons\mathrm{2H}^{+} +\mathrm{CO}_{\mathrm{3}} ^{−\mathrm{2}} \\ $$$$\:\:\mathrm{K}_{\mathrm{2}} =\frac{\left[\mathrm{2H}^{+} \right]^{\mathrm{2}} \left[\mathrm{CO}_{\mathrm{3}} ^{−\mathrm{2}} \right]}{\left[\mathrm{HCO}_{\mathrm{3}} ^{−} \right]\left[\mathrm{H}^{+} \right]}\:\rightarrow\mathrm{K}_{\mathrm{2}} =\mathrm{10}^{−\mathrm{9},\mathrm{28}} \\ $$$$ \\ $$