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M Common HW 0 (LAST X C fi ezto.mheducation.com /hm.tpx Apps P Pandora O EMAIL Moodle Top Hat W whentowork N Netflix webadvisor BO M CHEM Connect CALC chem rent E EMBE West Virginia Histor... E sign In: Discover! 03&E1; Common HW #10 (LAST!) instructions help Ea Save & Exit Submit Question 16 (of 16 16 6.25 points 1 out of 3 attempts Be sure to answer all parts Assistance Check My Work What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 0.400 L of a buffer with Question Help the same pH as a buffer made from 475 mL of 0.200M benzoic acid and 25 mL of 2.00 M Report a Problem NaOH? Volume of HCOOH L HCOOH Volume of NaOH L NaOH 2:32 PM /30/2015

Explanation / Answer

we know that

moles = molarity x volume (L)

so

moles of C6H5COOH = 0.2 x 0.475 = 0.095

moles of NaOH = 2 x 0.025 = 0.05

now

C6H5COOH + NaOH --> C6H5COONa + H20

moles of C6H5COOH reacted = moles of NaoH added = 0.05

moles of C6H5COONa formed = moles of NaOH added = 0.05

now

moles of C6H5COOH remaining = 0.095 - 0.05 = 0.045

now

pH = pKa + log [ C6H5COONa / C6H5COOH ]

so

pH = 4.202 + log [ 0.05 / 0.045]

pH = 4.2477

now

for HCOOH

pH = pKa + log [ HC00Na / HCOOH ]

4.24777 = 3.77 + log [ HCOONa / HCOOH ]

[HCOONa / HCOOH ] = 3

[HCOONa ] = 3 [HCOOH]

as the final volume is same for both

ratio of concentrations = ratio of moles

so

moles of HCOONa = 3 x moles of HCOOH

now

let the volume of NaOH be V

then volume of HCOOH = 0.4 - V

now

moles = conc x volume

so

moles of NaoH = 2 x V = 2V

moles of HCOOH = 0.2 x ( 0.4 - V) = 0.08 - 0.2V

now

HCOOH + NaOH ---> HCOONa + H20

moles of HCOOH reacted = moles of NaOH added = 2V

moles of HCOONa formed = moles of NaOH added = 2V

moles of HCOOH remaining = 0.08 - 0.2V - 2V

moles of HCOOH remainign = 0.08 - 2.2 V

now

moles of HCOONa = 3 x moles of HCOOH

so

2V = 3 x ( 0.08 - 2.2V)

V = 0.028

so

volume of NaOH = 0.028 L

volume of HCOOH = 0.4 - V = 0.4 - 0.028 = 0.012 L

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