Question

Mohon bantuanya kakak2 sekalian, mohon djawab dgn benar. Terimakasih ​

Answer 1

1) NaOH + CH3COOH ->> NaCH3COO + H2O

m 20 50 - -

r 20 20 20 20

s - 30 20 20

Asam lemah dan garam yang tersisa ->> Buffer asam

[H⁺] = Ka [Al]/[g]v

[H⁺] = 10⁻⁵ [30]/20x1

[H⁺] = 1,5 x 10⁻⁵

PH = 5- log 1,5

2) HNO3 + NH4OH ->> NH4NO3 + H2O

m 8 32 - -

r 8 8 8 8

s - 16 8 8

Basa lemah dan garam yang tersisa ->> Buffer basa

[OH⁻] = Kb [Bl]/[g] v

[OH⁻] = 10⁻⁵ [16]/8 x 1

[OH⁻] = 2 x 10⁻⁵

POH = 5-log 2

PH = 9 + log 2

3) PH akhir = 2 x PH awal = 2 x3 = 6

->> [H⁺] = 10⁻⁶

[H⁺] = Ka [Al]/[g] x V

10⁻⁶ = 10⁻⁵ [10²]/[g]xV

10⁻⁶ = 10⁻³/[g]

[g] = 10⁻³/10⁻⁶ = 10³ mmol

Mr natrium asetat = 72
massa = nMr = 72 x 10³ mg atau 72 gram