Tolong di jawab pake caranya kak​

Jawaban:

Bagian A

HNO3 –> [H+] + [NO3-]

[H+] = Ma x a

[H+] = 0,02 x 1

[H+] = 0,02 atau 2×10^-2

pH = -log [H+]

pH = -log 2×10^-2

pH = 2 – log 2

pH = 2 – 0,3

pH = 1,7

Bagian B

HI –> [H+] + [I-]

[H+] = Ma x a

[H+] = 0,01 x 1

[H+] = 0,01 atau 1×10^-2

pH = -log [H+]

pH = -log 1×10^-2

pH = 2 – log 1

pH = 2 – 0

pH = 2

Bagian C

KOH –> [K+] + [OH-]

[OH-] = Mb x b

[OH-] = 0,03 x 1

[OH-] = 0,03 atau 3×10^-2

pOH = -log [OH-]

pOH = -log 3×10^-2

pOH = 2 – log 3

pOH = 2 – 0,47

pOH = 1,53

pH = pKW – pOH

pH = 14 – 1,53

pH = 12,47

Bagian D

Ca(OH)2 –> [Ca2+] + [2OH-]

[OH-] = Mb x b

[OH-] = 0,002 x 2

[OH-] = 0,004 atau 4×10^-3

pOH = -log [OH-]

pOH = -log 4×10^-3

pOH = 3 – log 4

pOH = 3 – 0,6

pOH = 2,4

pH = pKW – pOH

pH = 14 – 2,4

pH = 11,6