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25 ml of 0.02M means 0.02 x251000 moles are pres.pdf

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anonamobilesp

25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95 Solution 25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95.

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25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95 Solution 25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95

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25 ml of 0.02M means 0.02 x251000 moles are pres.pdf

  • 1. 25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95 Solution 25 ml of 0.02M means 0.02 x25/1000 moles are present = 5 x10^-4 moles SH----- >S- +H+ ka=[H+][S-]/[SH] 2.1 x10^-12 = x^2/(5x10^-4 -x) 10.5 x10^-16 =x^2 x=[H+]=3.24 x10^-8 H2O will have 10^-7 of [H+] total [H+]=10^-8(1+0.1) pH=-log(1.1 x10^-8) =8-log(1.1) =7.95