# Contoh Soal Asam Basa : Contoh Soal No 11-20

11. Perhatikan Keterangan Di Bawah !
a) H2SO4 0,02 M

b) HNO3 0,002 M

c) CH3COOH 0,1 M (Ka =1,8 x 10-5)

d) HNO2 0,2 M ( Ka = 7,2 x 10-4)

Bagaimanakah urutan asam berdasarkan kenaikan harga pH ?

Jawaban : a, d, b, c.

Pembahasan :

[ H+] H2SO4 = indeks asam x M

= 2 x 0,02 M

= 0,04 M

= 4 x 10-2 M

pH H2SO4 = - log [H+]

= - log 4 x 10-2

= 2 – log 4

= 1,39794

[ H+] HNO3 = indeks asam x M

= 1 x 0,002 M

= 0,002 M

= 2 x 10-3 M

pH HNO3 = - log [H+]

= - log 2 x 10-3

= 3 – log 2

= 2,69897

[ H+] CH3COOH = √ Ka x Ma

= √ 1,8 x 10-5 x 0,1 M

= √ 1,8 x 10-6

= 1,34 x 10-3

pH CH3COOH = - log [H+]

= - log 1,341640787 x 10-3

= 3 – log 1,341640787

= 2,8724

[ H+] HNO2 = √ Ka x Ma

= √ 7,2 x 10-4 x 0,2 M

= √ 1,44 x 10-4 x 0,2 M

= 1,2 x 10-2 M

pH HNO2 = - log [H+]

= - log 1,2 x 10-2

= 2 – log 1,2

= 1,9208

12. Perhatikan slide!

a) Ba(OH)2 0,05 M

b) C5H5N 0,1 M (Kb = 1,7 x 10-9)

c) NH3 0,4 M (Kb = 1,8 x 10-5)

d) Mg(OH)2 0,2 M

Bagaimanakah urutan basa berdasarkan penurunan harga pH ?

Jawaban : d, a, c, b.

Pembahasan :

[ OH-] Ba(OH)2 = indeks Basa x M

= 2 x 0,05 M

= 0,1 M

= 10-1 M

pOH Ba(OH)2 = - log [OH-]

= - log 10-1

= 1

pH Ba(OH)2 = 14 – 1 = 13

[ OH-] C5H5N = √ Kb x Mb

= √ 1,7 x 10-9 x 0,1

= √ 1,7 x 10-10

= 1,303840481 x 10-5 M

pOH C5H5N = - log [OH-]

= - log 1,303840481 x 10-5

= 5 – log 1,303840481

pH C5H5N = 14 – (5 – log 1,303840481)

= 9 + log 1,303840481

= 9,1152

[ OH-] NH3 = √ Kb x Mb

= √ 1,8 x 10-5 x 0,4

= √ 7,2 x 10-6

= 2,683281573 x 10-3 M

pOH NH3 = - log [OH-]

= - log 2,683281573 x 10-3

= 3 – log 2,683281573

pH NH3 = 14 – (3 – log 2,683281573)

= 11 + log 2,683281573

= 11,4286

[ OH-] Mg(OH)2= indeks Basa x M

= 2 x 0,2 M

= 0,4 M

= 4 x 10-1 M

pOH Mg(OH)2= - log [OH-]

= - log 4 x 10-1

= 1- log 4

pH Mg(OH)2 = 14 – (1- log 4)

= 13 + log 4

= 13,6021

13. Perhatikan slide!

a) HNO3 0,001 M

b) HCl 0,0001 M

c) HCOOH 0,01 M (Ka = 1,7 x 10-4)

d) C6H5COOH 0,2 M ( Ka = 6,5 x 10-5)

Bagaimanakah urutan asam berdasarkan penurunan harga pH ?

Jawaban : b, a, c, d.

Pembahasan :

[ H+] HNO3 = indeks asam x M

= 1 x 0,001 M

= 1 x 10-3 M

pH HNO3 = - log [H+]

= - log 1 x 10-3

= 3

[ H+] HCl = indeks asam x M

= 1 x 0,0001 M

= 1 x 10-4 M

pH HCl = - log [H+]

= - log 1 x 10-4

= 4

[ H+] HCOOH = √ Ka x Ma

= √ 1,7 x 10-4 x 0,01

= √ 1,7 x 10-6

= 1,303840481 x 10-3 M

pH HCOOH = - log [H+]

= - log 1,303840481 x 10-3

= 3 – log 1,303840481

= 2,884775539

[ H+] C6H5COOH = √ Ka x Ma

= √ 6,5 x 10-5 x 0,2 M

= √ 13 x 10-6

= 3,6056 x 10-3 M

pH C6H5COOH = - log [H +]

= - log 3,6056 x 10-3

= 3 – log 3,6056

= 2,4430

14. Perhatikan slide!

a) C5H5N 0,3 M (Kb = 1,7 x 10-9)

b) C6H5NH2 0,04 M ((Kb = 7,4 x10 -10)

c) Ca(OH)2 0,002 M

d) NaOH 0,009 M

Bagaimanakah urutan basa berdasarkan kenaikan harga pH ?

Jawaban : b, a, c, d.

Pembahasan :

[ OH-] C5H5N = √ Kb x Mb

= √ 1,7 x 10-9 x 0,3= 5,1 x 10-10

= 2,258317958 x 10-5 M

pOH C5H5N = - log [OH-]

= - log 2,258317958 x 10-5

= 5 – log 2,258317958

pH C5H5N = 14 – (5 – log 2,258317958)

= 9 + log 2,258317958 = 9,3538

[ OH-] C6H5NH2 = √ Kb x Mb

= √( 7,4 x 10-10 x 0,04)

= √ 29,6 x 10-12

= 5,4405888203 x 10-6 M

pOH C6H5NH2 = - log [OH -]

= - log 5,4405888203 x10-6

= 6 – log 5,4405888203

pH C6H5NH2 = 14 – (6 – log 5,4405888203)

= 8 + log 5,4405888203

= 8,7356

[ OH-] Ca(OH)2= indeks Basa x M

= 2 x 0,002 M

= 0,004 M

= 4 x 10-3 M

pOH Ca(OH)2= - log [OH-]

= - log 4 x 10-3

= 3 - log 4

pH Ca(OH)2 = 14 – (3 - log 4)

= 11 + log 4

= 11,6021

[ OH-] NaOH = indeks Basa x M

= 1 x 0,009 M

= 9 x 10-3 M

pOH NaOH = - log [OH-]

= - log 9 x 10-3

= 3- log 9

pH NaOH = 14 – (3 - log 9)

= 11 + log 9

= 11,9542

15. Berapakah konsentrasi larutan anilin (C6H5NH2) dengan pH = 9 – log 5 ? (diketahui Kb = 7,4 x 10 -10)

Jawaban : 5,4054 x 10- 3 M

Pembahasan :

pH = 9 – log 5 = 8 + log 2

pOH = 14 – (8 + log 2)

pOH = 6 – log 2

[ OH-] = 2 x 10-6 M

[ OH-] = √ Kb x Mb

[ OH-]2 = Kb x Mb

(2 x 10-6)2 = 7,4 x 10 -10 x Mb

Mb = 4 x 10-12/ 7,4 x 10-10

Mb =0,54054 x 10-2 M = 5,4054 x 10-3 M

16. Berapakah konsentrasi larutan asam asetat dengan pH= 2 + log 5 ? (diketahui Ka = 1,8 x 10-5).

Jawaban : 2 ,2222 x 10-1 M

Pembahasan :

pH = 2 + log 5 = 3 – log 2

[ H+] = 2 x 10-3 M

[ H+] = √ Ka x Ma

[ H+]2 = Ka x Ma(2 x 10-3)2 = 1,8 x 10 -5 x Ma

Ma = 4 x10-6 / 1,8 x 10-5

Ma = 2 ,2222 x 10-1 M

17. berapakah harga Ka asam lemah HA 0,02 M dengan pH = 4 – log 6 ?

Jawaban : 1,8 x 10- 5

Pembahasan :

pH = 4 – log 6

[ H+] = 6 x 10-4 M

[ H+] = √ Ka x Ma

[ H+]2 = Ka x MA

(6 x 10-4)2 = Ka x 2 x 10-2

Ka = 36 x 10-5 / 2 x 10-1

Ka = 18 x 10-6 =1,8 x 10-5

18. Berapakah tetapan basa MOH 0,135 M dengan pH= 9 – log 1?

Jawaban : 7,4074 x 10- 10

Pembahasan :

pH = 9 – log 1 = 9

pOH = 14 – 9

pOH = 5

[ OH-] = 10-5 M

[ OH-] = √ Kb x Mb

[ OH-]2=  Kb x Mb

(10-5)2 = Kb x 0,135

Kb = 10-10/ 0,135

Kb = 7,4074 x 10- 10

19. berapakah konsentrasi larutan NH4OH dengan pH = 10 + log 3 ? (Kb NH4OH = 1,8 x 10-5.

Jawaban : 5 x 10-3 M

Pembahasan :

pH = 10 + log 3

pOH = 14 – (10 + log 3)

pOH = 4 – log 3

[ OH-] = 3 x 10-4 M

[ OH-] = 1,8 x 10-5 x Mb

[ OH-]2 = Kb x Mb

(3 x 10-4)2 = 1,8 x 10 -5 x Mb

Mb = 9 x10-8 / 1,8 x 10-5

Mb = 5 x 10-3 M

20. Berapakah harga pH larutan basa lemah LOH 0,025 M ? (Kb = 1,7 x 10 -9).

Jawaban : 8,81425

Pembahasan :

[ OH-] = √ Kb x Mb

[ OH-] = √ ( 25 X 10-3 ) x ( 1,7 x 10-9 )

[ OH-] = √ 42,5 x 10-12

[ OH-] = 6,519202405 x 10-6 M

pOH = 6 – log 6,519202405

pH = 14 - pOH

pH = 14 – (6 – log 6,519202405)

= 8 + log 6,519202405

= 8,8142

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