Question

Calculate the ph for each case in the titration of 50.0 ml of 0.150 m hclo(aq) with 0.150 m koh(aq). use the ionization constant for hclo

Answer 1

Answer;

pH =7.58

Explanation;

The KOH will react with the HClO to produce KClO. You will have a solution containing unreacted HClO and KOH. This is a buffer solution.  

Equation:  

HClO + KOH → KClO + H2O  

HClO reacts with KOH in 1:1 molar ratio  

Mol HClO in 50mL of 0.150M solution = 50/1000*0.150 = 0.0075 mol HClO  

Mol KOH in 30mL of 0.150M solution = 30/1000*0.150 = 0.0045 mol KOH  

These react to produce 0.0045 mol KClO and there is 0.0030 mol HClO unreacted  

Volume of solution = 50mL + 30mL = 80mL = 0.080L  

Molarity of HClO in solution = 0.0030/0.080 = 0.0375M  

Molarity of KClO in solution = 0.0045/0.080 = 0.0562M  

Using the Henderson-Hasselbalch equation we can calculate the pH;  

pKa HClO = -log ( 4.0*10^-8) = 7.40  

pH = pKa + log ([KClO]/[HClO])  

pH = 7.40 + log( 0.0562/0.0375)  

pH = 7.40 + log 1.50  

pH = 7.40+ 0.18  

pH = 7.58

Answer 2

PH = 7.58

Further Explanation

KOH will interact with HClO to provide KClO. you may have an answer that contains unreacted HClO and KOH. this is often a solution.

Equation:

HClO + KOH → KClO + H2O

HClO reacts with KOH in a very 1: 1 molar ratio

Mole of acid in 50 ml of an answer of 0.150 m = 50/1000 * 0,150 = 0.0075 mole of acid

Mole of KOH in 30 ml of 150 M solution = 30/1000 * 0,150 = 0.0045 mole of KOH

This reacts to provide 0.0045 moles of KClO and there are 0.0030 moles of HClO unreacted

The volume of the answer = 50 ml + 30 ml = 80 ml = 0.080 l

Deposition of acid in solution = 0.0030 / 0.080 = 0.0375M

KClO deposit in solution = 0.0045 / 0.080 = 0.0562 m

Using the Henderson-Hasselbalch equation, we will calculate pH;

pKa HClO = -log (4.0 * 10 ^ -8) = 7.40

PH = pKa + log ([KClO] / [HClO])

PH = 7.40 + notes (0.0562 / 0.0375)

PH = 7.40 + 1.50 notes

PH = 7.40+ 0.18

PH = 7.58

Learn More

Mole of Acid  brainly.com/question/9465562

HClO  brainly.com/question/12355703

Details

Grade: College

Subject: Chemistry

Keyword: mole, acid, HClO