Question

Mohon bantuan beserta penjelasannya ya hehe
1. pH dr suatu larutan yg konsentrasi ion H+ sebesar 10^-4 M
2. pH dr suatu larutan yg konsentrasi ion H+ sebesar 2x10^-4 M (log 2=0,3)
3. pH larutan A= 3. Besar konsentrasi ion H+ dlm larutan itu
4. pH larutan A=2,7. besar konsentrasi ion H+ dlm larutan tsb (log 2 = 0,3)
5. larutan A ber pH 4,5 . pOH dalam latutan tsb
6. pOH= 11,2. Brp pH nya
7. tetapan asam HA 10^-5 , pH larutan HA 0,1 M
8. pH larutan H2SO4 0,005 M
9. pH larutan NH3 0,1 M jika Kb=10^-3
10. Kb NH3=2x10^-5 , pH NH3 0,2 M?
terimakasih banyak

Answer 1

1. pH= -log [H+] = -log (10^-4) = 4
2. pH= - log [H+]= -log 2x10^-4= 4-log 2= 4-0,3 = 3,7
3. pH= 3 berarti  [H+] = 10^-3
4. pH= 2,7 berarti [H+] = 10^-2,7
5. pOH= 14- pH= 14-4,5= 9,5
6. pH= 14-pOH=14-11,2= 2,8
7. [H+]= (akar)Ka [HA] = (akar) 10^-5 (0,1)= (akar) 10^-6= 10^-3 , pH= -log [H+] = -log (10^-3) = 3
8. [H+]= x[H2SO4] = 2 (0,005) = 0,01= 10^-2 , pH= -log [H+] = -log (10^-2) = 2
9. [OH-]= (akar) Kb [NH3] = (akar) 10^-3 (0,1) = (akar) 10^-4= 10^-2, pOH= -log [OH-]= -log 10^-2 = 2 . pH= 14-pOH=14-2= 12
10. [OH-]= (akar) Kb [NH3]= (akar) 2x10^-5 (0,2) = (akar) 4x10^-6= 2x10^-3 . pOH= -log [OH-] = -log (2x10^-3) = 3-log2 , pH= 14-pOH= 14- (3-log2)= 11+log2=11, 30103