[H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2.pdf

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[H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2: Ka2 = [H+][SO42-]/[HSO4-] = 1.3X10^-2 From the starting condition, let x = [SO42-]. Then, [H+] = 0.14+ x, and [HSO4-] = 0.14 - x. now the value in Ka2 expression: 1.3X10^-2 = (0.14 +x) (x) / (0.14-x) 1.82 X10^-3 - 1.3X10^-2x = 0.14 x + x^2 x^2 + 0.153 x – 1.82 X10^-3 = 0 solve the above equation we will get; x = 0.0111 thus; x = [SO42-]=0.0111 Then, [H+] = 0.14 + 0.0111 = 0.1511 , and [HSO4-] = 0.14 – x = 0.14- 0.0111= 0.1289 Solution [H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2: Ka2 = [H+][SO42-]/[HSO4-] = 1.3X10^-2 From the starting condition, let x = [SO42-]. Then, [H+] = 0.14+ x, and [HSO4-] = 0.14 - x. now the value in Ka2 expression: 1.3X10^-2 = (0.14 +x) (x) / (0.14-x) 1.82 X10^-3 - 1.3X10^-2x = 0.14 x + x^2 x^2 + 0.153 x – 1.82 X10^-3 = 0 solve the above equation we will get; x = 0.0111 thus; x = [SO42-]=0.0111 Then, [H+] = 0.14 + 0.0111 = 0.1511 , and [HSO4-] = 0.14 – x = 0.14- 0.0111= 0.1289.

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1.82 X10^-3 - 1.3X10^-2x = 0.14 x + x^2
x^2 + 0.153 x – 1.82 X10^-3 = 0
solve the above equation we will get;
x = 0.0111
thus;
x = [SO42-]=0.0111
Then, [H+] = 0.14 + 0.0111 = 0.1511
, and [HSO4-] = 0.14 – x = 0.14- 0.0111= 0.1289

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[H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2.pdf

1. [H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2: Ka2 = [H+][SO42-]/[HSO4-] = 1.3X10^-2 From the starting condition, let x = [SO42-]. Then, [H+] = 0.14+ x, and [HSO4-] = 0.14 - x. now the value in Ka2 expression: 1.3X10^-2 = (0.14 +x) (x) / (0.14-x) 1.82 X10^-3 - 1.3X10^-2x = 0.14 x + x^2 x^2 + 0.153 x – 1.82 X10^-3 = 0 solve the above equation we will get; x = 0.0111 thus; x = [SO42-]=0.0111 Then, [H+] = 0.14 + 0.0111 = 0.1511 , and [HSO4-] = 0.14 – x = 0.14- 0.0111= 0.1289 Solution [H2SO4] = 0 because H2SO4 is a strong acid. first consider Ka2: Ka2 = [H+][SO42-]/[HSO4-] = 1.3X10^-2 From the starting condition, let x = [SO42-]. Then, [H+] = 0.14+ x, and [HSO4-] = 0.14 - x. now the value in Ka2 expression: 1.3X10^-2 = (0.14 +x) (x) / (0.14-x)

2. 1.82 X10^-3 - 1.3X10^-2x = 0.14 x + x^2 x^2 + 0.153 x – 1.82 X10^-3 = 0 solve the above equation we will get; x = 0.0111 thus; x = [SO42-]=0.0111 Then, [H+] = 0.14 + 0.0111 = 0.1511 , and [HSO4-] = 0.14 – x = 0.14- 0.0111= 0.1289

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