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nasty-shy [4]
4 years ago
12

(b) what is the ph of a solution containing 0.03m benzoic acid (pka 3.19) and 0.02m nabenzoate?

Chemistry
1 answer:
Makovka662 [10]4 years ago
5 0
Answer is: pH of solution is 3,02.
c(benzoic acid) = 0,03 mol/dm³.
c(sodium benzoate) = 0,02 mol/dm³.
pKa(benzoik acid) = 3,19.
pH = ?
Henderson–Hasselbalch equation for buffers: pH = pKa + log(cs/ck)
pH = 3,19 + log(0,02/0,03)
pH = 3,19 - 0,176 
pH = 3,02.
cs - concentration of acid.

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cestrela7 [59]

Answer:

Ka = 0.1815

Explanation:

Chromic acid

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Ka = ?

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Step 1 - Set up Initial, Change, Equilibrium table;

H2CrO4 ⇄  H+   +   HCrO4−

Initial - 0.078M 0   0

Change : -x    +x       +x

Equilibrium : 0.078-x    x      x

Step 2- Write Ka as Ratio of Conjugate Base to Acid

The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].

Step 3 - Plug in Values from the Table

Ka = x * x / 0.078-x

Step 4 - Note that x is Related to pH and Calculate Ka

[H+] = 10^-pH.

Since x = [H+] and you know the pH of the solution,

you can write x = 10^-1.23.

It is now possible to find a numerical value for Ka.

Ka =  (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156

Ka = 0.1815

4 0
3 years ago
You have a solution with an unknown concentration of hcl, in order to determine the concentration of hcl you titrate 35.00ml of
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