Hitunglah ph larutan A. Larutan H= so4 0,1m B.100 ml larutan ba(oH)2 = 5x10^-4m C.larutan HBRO o,1 m ka= 4x10^-9 D. Larutan NH4 OH 0,1 m kb= 10^-5 tolong bantu

A. HSo4=0,1
[H+] = a.Ma
=1. 0,1
=0,1 =1x10^-1
pH= - log [H+]
=-log[1x10^-1]
=1-log 1
B. Ba(OH)2 = 5x10^-4 m
[OH-] = b.Mb
= 2 . 5x10^-4
= 10x10^-4
pOH = -log [OH-]
= - log 10x10^-4
=4-log 10
ph= 14-poh
=14- (4 - log 10)
=10+log 10
c. [OH-] =
[tex] \sqrt{kw \div ka \times m} [/tex]
[tex] \sqrt{10 {}^{ -14} \div 4 \times 10 {}^{ - 9} \times 10 {}^{ - 1} } [/tex]
[tex] \sqrt{ \frac{5}{2} \times 10 {}^{ - 6} \times 10 {}^{ - 1} } [/tex]
[tex] \sqrt{2.5 \times 10 {}^{ - 7} } [/tex]
[tex]1.8 \times 10 {}^{ - 5.5} [/tex]
poh=-log [oh]
=-log 1.8 x 10^-5.5
=5.5-log 1.8
ph= 14-poh
= 14- (5.5-log1.8)
=8.5+log1.8
d. [H+]=
[tex] \sqrt{kw \div kb \times m} [/tex]
[tex] \sqrt{10 {}^{ - 14} \div 10 {}^{ - 5 } \times 10 {}^{ - 1} } [/tex]
[tex] \sqrt{10 {}^{ - 10} } [/tex]
[tex]10 {}^{ - 5} [/tex]

pH = -log [H+]
= -log 10^-5
=5- log 10