Solution ñ0,0 ml ML O1015 M 01015 M of 十he ution by equafion)

Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter6: Solutions And Colloids
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Problem 6.104P
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Solution ñ0,0 ml
ML
O1015 M
01015 M
of
十he
ution
by
equafion)
Transcribed Image Text:Solution ñ0,0 ml ML O1015 M 01015 M of 十he ution by equafion)
Expert Solution
Step 1

Part a:

Number of moles of HCl solution = 0.015 M × 0.05 L = 0.00075 mol

Number of moles of NaOH solution = 0.015 M × 0.01 L = 0.00015 mol

0.00015 mol of NaOH neutralizes 0.00015 mol of HCl and forms 0.00015 mol of NaCl. The remaining number of moles of HCl is equal to 0.00075 – 0.00015 mol = 0.0006 mol. The pH of the solution is obtained due to 0.0006 mol of HCl since NaCl is a salt of strong acid and strong base and such salts do not undergo hydrolysis. The neutralization reaction between NaOH and HCl can be shown by the following balanced chemical equation:

Chemistry homework question answer, step 1, image 1

Step 2

Since HCl is a strong acid, it completely dissociates into hydrogen ions and chloride ions. The total number of moles of hydrogen ions released from 0.0006 mol of HCl is equal to 0.0006 mol and the total volume of the solution is equal to 0.05 L + 0.01 L = 0.06 L. The pH of the solution can be calculated as follows-

Chemistry homework question answer, step 2, image 1

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