Contoh Soal Asam Basa : Teladan 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 menurut 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 menurut 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 menurut 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 menurut 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 Sumber http://www.panduankimia.net
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 menurut 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 menurut 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 menurut 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 menurut 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 Sumber http://www.panduankimia.net
Post a Comment for "Contoh Soal Asam Basa : Teladan Soal No 11-20"