sood.cr\'s Me\'\". Long questions: (28 points) Show your detailed calculation an
ID: 694904 • Letter: S
Question
sood.cr's Me'". Long questions: (28 points) Show your detailed calculation and/or explanation to receive 1 mypochlorous acid Hoci is a weak acid with K..40 x 10, 300 mL of 0AHOO is (a) (2pt) Determine the number of mole of HOci in the 30.0-mi solution of 0.40 (b) (2pt) Determine the number of mole of NaOH in a 48.0-ml solution ofo (c) (8pt) Determine the pH of the after 30.0 NaOH. Show your detailed calculation. (d) (4pt) If we add 1 is titrated with 025-M NaDH 2S M NaOH atonture at equilibrium atter 30.0 mi of 0.A0 14 HOCi is mised with s 0 m o 0 2 5.0 ml of 0.25-M NaOH into 30.0 mL of 0.40-M HOCI, then the final solution should have the on value (greater than, equal to, or less than) pK,? Explain your choice (e) (3pt) Which of the following acids is a stronger acid: (hypochirur choice a) HCIO acid woci or perchlonic acid wOD? Exglain your d) none of above b) HOCI c) same for bothExplanation / Answer
a) use the concentration formula to find out moles of HOCl
molarity = numbers of moles of solute / volution of solution (L)
moles of HOCl = molarity x volume
moles of HOCl = 0.4 M x 0.030 L = 0.012 moles
b) same for NaOH
moles of NaOH = molarity x volume
moles of NaOH = 0.25 M x 0.0480 L = 0.012 moles
c) pH of titration of equal moles of HOCl and NaOH. this point called the equivalence point - and calculate the pH at the equivalence point.
pH = pKa + log (base / acid)
we need ka so ,
ka = (1.0x10-14 / 4.0x10-8)
ka = 2.5x10-7
pH = -log(2.5x10-7) + log( 1 )
pH = 6.60
d) find numbers of moles of HOCl = 0.012 moles
numbers of moles of NaOH = 0.015 L x 0.25 M = 0.00375 moles
after titration of NaOH and HOCl
HOCl remain excess amount in solution so
pH = pKa + log (base / acid )
pH = 6
pH < pKa
e) oxygens bonded to the central atom increases, the oxidation number of the central atom increases causing a weakening of the O-H bond strength and an increase in the acidity so, Most acidic: HClO4 > HOCl.
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