Sebanyak 10 mL larutan asam asetat dengan pH=3 dicampurkan dengan 90 mL air. Berapakah pH larutan yang sekarang ?
a. 2,5
b. 3
c. 3,5
d. 4
e. 4,5
tolong dibantu ya , kalo bisa dengan cara biar jelas hehee :)
a. 2,5
b. 3
c. 3,5
d. 4
e. 4,5
tolong dibantu ya , kalo bisa dengan cara biar jelas hehee :)
rirismaulidina
![PH=3=>[H^+]=10^-^3~M \\ M1V1=M2V2 \\ M2= \frac{M1V1}{V2} \\ M2= \frac{(10^-^3)(10)}{(90+10)} \\ M2= \frac{10^-^2}{100} \\ M2=10^-^4 \\ H^+=10^-^4~M \\ PH=-~log~10^-^4 \\ PH=4~~(D)](https://tex.z-dn.net/?f=PH%3D3%3D%3E%5BH%5E%2B%5D%3D10%5E-%5E3%7EM+%5C%5C+M1V1%3DM2V2+%5C%5C+M2%3D+%5Cfrac%7BM1V1%7D%7BV2%7D++%5C%5C+M2%3D+%5Cfrac%7B%2810%5E-%5E3%29%2810%29%7D%7B%2890%2B10%29%7D++%5C%5C+M2%3D+%5Cfrac%7B10%5E-%5E2%7D%7B100%7D++%5C%5C+M2%3D10%5E-%5E4+%5C%5C+H%5E%2B%3D10%5E-%5E4%7EM+%5C%5C+PH%3D-%7Elog%7E10%5E-%5E4+%5C%5C+PH%3D4%7E%7E%28D%29 "PH=3=>[H^+]=10^-^3~M \\ M1V1=M2V2 \\ M2= \frac{M1V1}{V2} \\ M2= \frac{(10^-^3)(10)}{(90+10)} \\ M2= \frac{10^-^2}{100} \\ M2=10^-^4 \\ H^+=10^-^4~M \\ PH=-~log~10^-^4 \\ PH=4~~(D)")
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