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THoneymoon x Ga Blackboard L x D bboard2010 x E-Mail Accesx E Sign into yo x C Chemistr q My GRE My x GILove Kanye x CHEM 1215 X C WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-2501029 Sapling Learning Sapling Learning CHEM 1215-General Chemistry I-Wickham Activities an My Assignment Resources 4/25/2016 11:55 PM 71.9/100 Gradebook O Assignment Information Attempts Score Print HCalculator Periodic Table Available From: 4/10/2016 06.00 PM 4/25/2016 11:55 PM Due Date: Question 5 of 8 Points Possible: 100 Map Dab General Chemistr University Science Books Grade Category: Graded Description You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (pKa E 4.20 and 0.140 M sodium b You can check your answers. You can view solutions when you complete or give up on any question. much of each solution should be mixed to prepare this buffer? You can keep trying to answer each question until you get it right or give up Number You lose 5% of the points available to each answer mL benzoic acid. in your question for each incorrect attempt at that O Help With This Topic Numb mL odium benzoate Web Help & Videos Technical Support and Bug Reports Previous 3 Give Up & View Solution Check Answer Next Exit careers partner prhag polly terms ofuse contact us m Cortana. Ask me anything Logout 11:44 PM n & 4/25/2016

Explanation / Answer

Answer: Given benzoic acid=0.100M and sodium benzoate 0.140M and pKa= 4.20

so as we know

pH = pKa + log [sodium benzoate]/[benzoic acid]

4 = 4.20 + log [sodium benzoate]/[benzoic acid]

-0.20 = log [sodium benzoate]/[benzoic acid]

apply anitlog

[sodium benzoate]/[benzoic acid]= 0.631

no. of moles of sodium benzoate / no. of moles of benzoic acid = 0.631

molarity = no. of moles of solute / volume of solutionin litres

So no. of moles of solute = molarity X volume of solution in litres

as the Buffer solution is of 100 ml or 0.1 L

let Volume of Benzoic acid be x L

then Sodium benzoate will be 0.1-x L

no. of moles of benzoic acid = 0.1 X x
no. of moles of Sodium benzoate = 0.140 X (0.1-x)

putting in above relation

no. of moles of Sodiumbenzoate / no. of moles of benzoic acid = 0.631
0.140(0.1-x) / 0.1x = 0.631

0.140(0.1-x) = 0.1x X 0.631 = 0.0631x

0.014 - 0.140x = 0.0631x
0.014 = 0.140x + 0.0631x = 0.2031x

x = 0.014/0.2031 = 0.0689 L

therefore x = volume of acid used = 0.0689 L or 68.9ml

0.1-x = volume of salt used = 0.1-0.0689 = 0.0311 L or 31.1ml

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