January 23, 2021, 10:08:48 PM
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### Topic: hydrolysis of sucrose  (Read 150 times)

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#### penovili

• Very New Member
• • Posts: 1
• Mole Snacks: +0/-0 ##### hydrolysis of sucrose
« on: January 15, 2021, 03:58:37 AM »
Hi guys,
So i got the K1 value for this reaction wich was done using HCL
Now i have to plot the graph of k1 vs H3O.
The pka of hcl is -7
what would be [H3O]

I know that [H3O]+ = [H]+

attempt :
pka= -logka= -7
logka= 7
ka=10^7
ka= x^2/ (3.0M - x) aprrox = x^2/3.0M
1.0 * 10^7 = x^2/3.0M

x= [H3O]

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1838
• Mole Snacks: +256/-12 ##### Re: hydrolysis of sucrose
« Reply #1 on: January 15, 2021, 06:33:22 AM »
You can't make the assumption 3.0 - x ≈ 3.0. That only works for weak acids where x is small. Your calculation would lead to x ≈ 103M - does that look sensible? pKa = -7 means it is a strong acid that is essentially fully ionised in water. So for 3.0 M HCl, you can take [H3O+] as 3.0 M.

What you could also do is call the concentration of undissociated HCl y; then [H3O+] = [Cl-] = 3.0 - y, and
Ka = (3.0 - y)2/y
and simplify by assuming 3.0 - y ≈ 3.0 (Check the validity of this assumption after calculating y.)