PPR - Endocrine University of Oklahoma saplinglearning.com saplinglearning.com O

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PPR - Endocrine University of Oklahoma saplinglearning.com saplinglearning.com O saplingleaming.com/ibiscms/mod/ibi5Nie w.php?id-3871 623 Sapling Learning Andrea Brisby Jump to... macmillan leaming Map Due Date: Points Po Grade Cat Descripti Policies: Solutions 2 Sapling Learning macmillon learning A 1.44 L buffer solution consists of 0.111 M butanoic acid and 0.270 M sodium butanoate. Calculate the pH of the solution following the addition of 0.062 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 10 Number 6 You can You can You have There is 9 eTextbo 10 Help Wi 12 Web He 13 0 2011-2017 Sapling Learning, Inc. about uscareersprivacy policy terms of use contact us help 946 PM Type here to search @ ND4») 10242017 ES

Explanation / Answer

mol of NaOH added = 0.062 mol

C3H7COOH will react with OH- to form C3H7COO-

Before Reaction:

mol of C3H7COO- = 0.27 M *1.44 L

mol of C3H7COO- = 0.3888 mol

mol of C3H7COOH = 0.111 M *1.44 L

mol of C3H7COOH = 0.1598 mol

after reaction,

mol of C3H7COO- = mol present initially + mol added

mol of C3H7COO- = (0.3888 + 0.062) mol

mol of C3H7COO- = 0.4508 mol

mol of C3H7COOH = mol present initially - mol added

mol of C3H7COOH = (0.1598 - 0.062) mol

mol of C3H7COOH = 0.0978 mol

Ka = 1.52*10^-5

pKa = - log (Ka)

= - log(1.52*10^-5)

= 4.8182

since volume is both in numerator and denominator, we can use mol instead of concentration

we have below equation to be used

pH = pKa + log {[conjugate base]/[acid]}

= 4.8182+ log {0.4508/0.0978}

= 5.48

Answer: 5.48

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