A 30 mL sample of 0.100 M HOBr is titrated using 0.100 NaOH. Ka= 2.5x10-9 for...

A 30 mL sample of 0.100 M HOBr is titrated using 0.100 NaOH. K_a= 2.5x10^-9 for HOBr.

a) Write a balanced net-ionic equation for the titration reaction.

b) Calculate the pH of the titration mixture at the equivalence point.

c) Would bromthymol blue be a suitable indicator for the titration?

Solutions

Expert Solution

part a

HOBr + NaOH -> NaOBr +H2O

part b

at equivalence point moles of acid and base ae equal, nut the salt formed is of weak acid and strong base. this salt will get hydrolyzed and pH will not be 7 but greater than 7 as it is a salt of weak acid and strong base.

moles of acid

= 2.5*10^-9 * .1 = [acid]^2

= 1.58 * 10^-5 * 30 = 0.00047 millimoles

[OH-]= sqrt(Kb * [OBr-])

=sqrt( 0.000004* 1.58 * 10^-5)

conc of OH- = 0.00000794

pOH= -log(0.00000794) = 5.1

pH = 14 - 5.1 = 8.9

partc

yes it is a suitable indicator for this because its color change happens in the region of 6 - 7.6 which is ideal for this case.

queen_honey_blossom answered 1 year ago

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