# This question on pH will shock you.

LET’S SEE YOUR STRENGHT ON pH CALCULATION.

QUESTION.

Calculate the pH of a solution of CH3COOH whose acid dissociation constant  value is given as 1.84×10^-5 and 250cm³ of the solution contains 12g of the acid.

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May 26, 2020 2:13 pm

Ka=1.84×10^-5
V=0.25dm^3
mass=12g
PH=?
Mole=m/mm
M=12g
Mole=0.2
Conc.=mole/vol
C=0.2/0.25
C=0.8mol/dm^3
PH=-log√ka.c
PH=-log√1.84×10^-5×0.8
PH=-log√1.472×10^-5
PH=-log3.84×10^-3
PH=2.42
From Benitha Benson

May 25, 2020 8:48 pm

Good day sir catalyst am Daniel Oriola by name
Sir I have not been attending your class and I am in your group please can I still cope

May 25, 2020 8:43 pm

Good day sir catalyst am Daniel Oriola by name
Sir I have not been attending your class and I am in your group please can I still cope

May 25, 2020 8:18 pm

I have forgoten to put my name 😜😜👆(Thomas)

May 25, 2020 8:16 pm

CH3COOH +H2O—->CH3COO^- and H3O^+
Ka = [CH3COOH]^- +
[H3O]^+/[CH3COOH
Ka=1.84*10^-5 CH3OOH=0.8M-x
The two products =x²
Where x = unresolved ions
:. 0.0000184=x²/0.8-x
Since x I negligible
X=√(0.0000184*0.8)
X=0.0038
pH =-log[H]^+
Where[H]^+= x = 0.0038
pH=-log(0.0038)
pH=2.42

May 25, 2020 6:24 pm

Isaac M
PH = -log√ka.C
C=mass conc/molar mass
Mass con =molarity ×m.mass
Molarity= mole×1000/vol
=0.2×1000/250
= 0.8molperdm³
Mass con.=0.8×60=48g/dm³
=C=48/60=0.8molperdm³
PH=-log√1.8×10^-5×0.8
=2.42

May 25, 2020 2:42 pm

PH=4.04

May 25, 2020 2:35 pm

Name:chinex

May 25, 2020 2:33 pm

ChiChi
CH3COOH » CH3COO^- + H^+
That means one mole of H^+ is produced
Ka = Acid Dissociation Constant
[H^+] = √{Ka × C}

Ka = 1.84 × 10^-5
C = Mass Conc. ÷ Molar Mass

12 ÷ (250÷1000) = 12 ÷ 0.25 = 48 ~ Mass Conc
C + 3H + C + O + O + H = 12+3+12+ 16+16+1 = 60g/Mol ~ Molar Conc

C = 48÷60 = 0.8

[H^+] = √{1.84 ×10^-5} × {0.8}
=√0.00001472
=0.0038366 or 3.8366 × 10^-3
pH = -Log10{H^+}
= -Log10{3.8366 × 10^-3}
pH = 2.4160534 or 2.42 (2 D.P)

May 25, 2020 2:32 pm

PH of CH3COOH=-LOG√[Ka.c]
C=concentration
C=n/v
n=mass/molar mass
n=12/60=0.2
Vol.=250cm³=0.25L
C=0.2/0.25=0.8
Ka=1.8×10*-5
PH=-log√[1.8×10*-5×0.8]
PH=4.84

May 25, 2020 2:26 pm

Danny

pH = pka + log ([anion]/[acid])

CH3cooh —-CH3COO^- + H^ +

To find the mole concentration of the anion and acid
We do

Ka = [CH3COO^-] [ H^+]
————————-
CH3COOH
Where ka is given as 1.84 ×10^-5
To find the mole concentration of CH3COOH the acid we do

Mol = mass
——-
Molar mass
=12/60 =0.2

Then mole concentration is mole/volume = 0.2 /0.25
=0.8 mol per dm3

1.84 ×10^-5 =[×][×]
——
0.8
X² =0.8 × 1.84 × 10^-5
X = 0.0037
Thus anion CH3CO0^- = 0.0037

Pka = -log(ka) that is ka = 1.84 ×10^-5
Pka= 4.74

pH = 4.74 + log([anion]/[acid])
pH = 4.74 +log (0.0037/0.0000144)
pH = 4.7 4 + 2.0(raised to nearest decimal)

pH = 6.74

May 25, 2020 2:14 pm

Incorrect.

May 25, 2020 1:53 pm

Chidocks

pH of CH3COOH = -2.303LogKa

Where Ka = acid dissociation constant

Ka = 1.84 × 10^-5

pH = -2.303Log(1.84 × 10^-5)
pH = 10.905 or 10.9 (1 D.P)

Ekene
May 26, 2020 9:41 pm

[H3o] = √ka[acid] Where [H3o] = hydrogen ion concentration Ka = dissociation constant [acid] = concentration of acid We are now looking for [acid] first. By going through these process 👇 250cm^3 = 12g Then 1000cm^3 = 3g. So therefore the concentration of the acid in g/dm^3 = 3g/dm^3 Then you have to convert to mol/dm^3, which is 👇 g/dm^3 divided by rel. Mol. Mass 3/60 = 0.05 mol/dm^3 Then you apply it to the equation [H3o] = √ka[acid]. To get the [H3o] [H3o] = √1.84×10^-5 × 0.05 [H3o]= √9.2×10^-7 [H3o]= 0.000959 Then we can now find the pH 👇… Read more »

Edmond
October 29, 2020 10:20 am

Edmond
October 29, 2020 10:21 am

How may I help?

January 22, 2021 5:08 pm

PH =-log√C×ka
C=m/mm
M=12g
Mm of CH3COOH =60M
C=12/60=0.2
PH=-log√0.2×1.8×10^-5
=2.72//

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