Larutan asam salisilat 0,200 m memiliki pH = 1,83. Hitung Ka dan pKa asam tersebut (pKa= -logKa) ?

1
[H^+] = √(K_(a ) [asam salisilat])
=√(K_(a ) [0,200])
pH= -log[H^+]
1,83= - log〖[H〗^+]
pH= - log 1,5 x 〖10〗^(-2)
= 2 - log 1,5
= 1,83
Jadi Ka=
[H^+] = √(K_(a ) [asam salisilat])
1,5 x 〖10〗^(-2) = √(K_(a ) 0,200)
1,5 x 〖10〗^(-2)= (K_(a ) 0,200)pangkat-1/2 □
〖0,015〗^2 =(K_(a ) 0,200)
0,000225= (K_(a ) 0,200)
〖 K〗_a = 0,000225/0,200
〖 K〗_a=0,001125 atau 0,001=〖10〗^(-3)
pKa= - log 〖10〗^(-3)
pKa= 3
[H^+] = √(K_(a ) [asam salisilat])
=√(K_(a ) [0,200])
pH= -log[H^+]
1,83= - log〖[H〗^+]
pH= - log 1,5 x 〖10〗^(-2)
= 2 - log 1,5
= 1,83
Jadi Ka=
[H^+] = √(K_(a ) [asam salisilat])
1,5 x 〖10〗^(-2) = √(K_(a ) 0,200)
1,5 x 〖10〗^(-2)= (K_(a ) 0,200)pangkat-1/2 □
〖0,015〗^2 =(K_(a ) 0,200)
0,000225= (K_(a ) 0,200)
〖 K〗_a = 0,000225/0,200
〖 K〗_a=0,001125 atau 0,001=〖10〗^(-3)
pKa= - log 〖10〗^(-3)
pKa= 3

Jawabanmu

2014-01-05T00:52:21+07:00
[H^+] = √(K_(a ) [asam salisilat])
=√(K_(a ) [0,200])
pH= -log[H^+]
1,83= - log〖[H〗^+]
pH= - log 1,5 x 〖10〗^(-2)
= 2 - log 1,5
= 1,83
Jadi Ka= 
[H^+] = √(K_(a ) [asam salisilat])
1,5 x 〖10〗^(-2) = √(K_(a ) 0,200)
1,5 x 〖10〗^(-2)= (K_(a ) 0,200)pangkat-1/2 □
〖0,015〗^2 =(K_(a ) 0,200)
0,000225= (K_(a ) 0,200)
〖 K〗_a = 0,000225/0,200
〖 K〗_a=0,001125 atau 0,001=〖10〗^(-3)
pKa= - log 〖10〗^(-3)
pKa= 3