Question

No.3 essai tolong bantu caranya dgn benar

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Answer 1

Mol Ba(OH)2 = 50x0,1 = 5 mmol

Mol HF = 100x0,2 = 20 mmol

2HF + Ba(OH)2 => BaF2 + 2H2O

20. 5. - -

10. 5. 5. 10

10. - 5. 10

a) sebelum ditambah HNO3

[H+] = (7x10^-4) (10/(5x2))

[H+] = 7x10^-4

pH = - log (7x10^-4) = 4 - log 7

b) setelah ditambah HNO3

Mol HNO3 = 10x0,2 = 2 mmol

2HNO3 + BaF2 => Ba(NO3)2 + 2HF

2. 5. - 10

2. 1. 1. 2

- 4 1. 12

[H+] = (7x10-4) (12/(4x2))

[H+] = 10,5x10-4

pH = - log (10,5x10^-4) = 4 - log 10,5