# 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

###### FREE Expert Solution

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

90% (392 ratings) ###### 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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