10) Diketahui :
V HCl = 50 mL = 0,05 L
pH HCl = 3
[H+] = 10^-3 M
[HCl] = [H+] : valensi asam = 10^-3 M : 1 = 10^-3 M
mol HCl = [HCl] x V HCl = 10^-3 x 0,05 L = 5 x 10^-5 mol
V NaOH = 50 mL = 0,05 L
pH NaOH = 12
pOH = 14 - pH = 14 - 12 = 2
[OH-] = 10^-2 M
[NaOH] = [OH-] : valensi basa = 10^-2 M : 1 = 10^-2 M
mol NaOH = [OH-] x V NaOH = 10^-2 x 0,05 = 5 x 10^-4 mol
jawab :
Pakai sistem Mula-mula Bereaksi Sisa (MBS)
HCl + NaOH ⇒ NaCl + H2O
5 x 10^-5 5 x 10^-4 - -
5 x 10^-5 5 x 10^-5 5 x 10^-5 5 x 10^-5
- 4,5 x 10^-4 5 x 10^-5 5 x 10^-5
jadi, karna NaOH masih bersisa 4,5 x 10^-4 mol, jadi pH NaOH sisa yang dihitung
*[NaOH] = mol NaOH : V Total = 4,5 x 10^-4 mol : 0,1 L = 4,5 x 10^-3 M
*[OH-] = valensi basa x [NaOH] = 1 x 4,5 x 10^-3 = 4,5 x 10^-3 M
*pOH = - log [OH-] = - log 4,5 x 10^-3 = 3 - log 4,5
*pH = 14 - pOH = 14 - ( 3 - log 4,5) = 11 + log 4,5 E
11) Diketahui :
*pH = 12 + log 5
*pOH = 14 - pH = 14 - ( 12 + log 5) = 2 - log 5
*[OH-] = 5 x 10^-2 M
*[NaOH] akhir = [OH-] : valensi basa = 5 x 10^-2 : 1 = 5 x 10^-2 M
#V HCl = 100 mL = 0,1 L
#[HCl] = 0,1 M
#mol HCl = [HCl] x V HCl = 0,1 M x 0,1 L = 0,01 mol
@[NaOH] mula2 = 0,2 M
@mol NaOH mula2 = [NaOH] x V NaOH = 0,2 x V = 0,2V mol
Jawab :
0,01 0,2V - -
0,01 0,01 0,01 0,01
- 0,01 - 0,2V 0,01 0,01
mol NaOH akhir = [NaOH] akhir x V Total
0,01 - 0,2V = 5 x 10^-2 x ( 0,1 + V )
0,01 - 0,2V = 5 x 10^-3 + 5 x 10^-2V
- 0,2V - 5 x 10^-2V = 5 x 10^-3 - 0,01
- 0,25V = - 5 x 10^-3
V = 0,02 Liter = 200 mL
Nb:
- Kalau ada yang gak paham, tanya aja
- Kalau misalnya editannya gak keliatan di jawabannya, mohon maaf yaa, saya bisanya ngetik langsung disini, kalau di kertas trus difoto, nanti blur, soalnya kamera laptop saya gak bagus kali
" Life is not a problem to be solved but a reality to be experienced! "
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10) Diketahui :
V HCl = 50 mL = 0,05 L
pH HCl = 3
[H+] = 10^-3 M
[HCl] = [H+] : valensi asam = 10^-3 M : 1 = 10^-3 M
mol HCl = [HCl] x V HCl = 10^-3 x 0,05 L = 5 x 10^-5 mol
V NaOH = 50 mL = 0,05 L
pH NaOH = 12
pOH = 14 - pH = 14 - 12 = 2
[OH-] = 10^-2 M
[NaOH] = [OH-] : valensi basa = 10^-2 M : 1 = 10^-2 M
mol NaOH = [OH-] x V NaOH = 10^-2 x 0,05 = 5 x 10^-4 mol
jawab :
Pakai sistem Mula-mula Bereaksi Sisa (MBS)
HCl + NaOH ⇒ NaCl + H2O
5 x 10^-5 5 x 10^-4 - -
5 x 10^-5 5 x 10^-5 5 x 10^-5 5 x 10^-5
- 4,5 x 10^-4 5 x 10^-5 5 x 10^-5
jadi, karna NaOH masih bersisa 4,5 x 10^-4 mol, jadi pH NaOH sisa yang dihitung
*[NaOH] = mol NaOH : V Total = 4,5 x 10^-4 mol : 0,1 L = 4,5 x 10^-3 M
*[OH-] = valensi basa x [NaOH] = 1 x 4,5 x 10^-3 = 4,5 x 10^-3 M
*pOH = - log [OH-] = - log 4,5 x 10^-3 = 3 - log 4,5
*pH = 14 - pOH = 14 - ( 3 - log 4,5) = 11 + log 4,5 E
11) Diketahui :
*pH = 12 + log 5
*pOH = 14 - pH = 14 - ( 12 + log 5) = 2 - log 5
*[OH-] = 5 x 10^-2 M
*[NaOH] akhir = [OH-] : valensi basa = 5 x 10^-2 : 1 = 5 x 10^-2 M
#V HCl = 100 mL = 0,1 L
#[HCl] = 0,1 M
#mol HCl = [HCl] x V HCl = 0,1 M x 0,1 L = 0,01 mol
@[NaOH] mula2 = 0,2 M
@mol NaOH mula2 = [NaOH] x V NaOH = 0,2 x V = 0,2V mol
Jawab :
HCl + NaOH ⇒ NaCl + H2O
0,01 0,2V - -
0,01 0,01 0,01 0,01
- 0,01 - 0,2V 0,01 0,01
mol NaOH akhir = [NaOH] akhir x V Total
0,01 - 0,2V = 5 x 10^-2 x ( 0,1 + V )
0,01 - 0,2V = 5 x 10^-3 + 5 x 10^-2V
- 0,2V - 5 x 10^-2V = 5 x 10^-3 - 0,01
- 0,25V = - 5 x 10^-3
V = 0,02 Liter = 200 mL
Nb:
- Kalau ada yang gak paham, tanya aja
- Kalau misalnya editannya gak keliatan di jawabannya, mohon maaf yaa, saya bisanya ngetik langsung disini, kalau di kertas trus difoto, nanti blur, soalnya kamera laptop saya gak bagus kali
#konsep stoikiometri
#XI semester 2
#kimiaschemistry