Problem: What is the pH of a 0.015 M solution of paraperiodic acid (H  5IO6) at 25°C?Ka1 = 2.8 × 10-2 and Ka2 = 5.3 × 10-9A) 1.69B) 1.55C) 2.04D) 1.97E) 1.67

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We are being asked to calculate the pH of the solution if the concentration of H5IOis 0.015 M

We're going to calculate the pH of the solution using the following steps: 

Step 1: Construct an ICE chart for the equilibrium reaction.
Step 2: Write the Ka expression.
Step 3: Calculate for the equilibrium concentration.
Step 4: Calculate pH.


Step 1: Construct an ICE chart for the equilibrium reaction.

Since we’re dealing with a weak acid and Ka is an equilibrium expression, we will have to create an ICE chart to determine the equilibrium concentration of each species:

H5IO6  weak acidproton donor
H2O → will act as the weak baseproton acceptor


We will use the first equilibrium reaction given:

       H5IO6  +  H2O (l) ⇋ H4IO6  (aq) + H3O(aq)               Ka1 = 2.8 x 10–2

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Problem Details

What is the pH of a 0.015 M solution of paraperiodic acid (H  5IO6) at 25°C?

Ka= 2.8 × 10-2 and Ka2 = 5.3 × 10-9

A) 1.69

B) 1.55

C) 2.04

D) 1.97

E) 1.67

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Our tutors have indicated that to solve this problem you will need to apply the Diprotic Acid concept. You can view video lessons to learn Diprotic Acid. Or if you need more Diprotic Acid practice, you can also practice Diprotic Acid practice problems.

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