Question

Consider the titration of a 35.0 mL sample of 0.170 M  HBr with 0.210 M KOH. Determine each of the following:

A) the initial pH [I calculated this and got pH =0.770]

B) the volume of added base required to reach the equivalence point

Express your answer in milliliters.

C) the pH at 10.2 mL of added base

Express your answer using three decimal places.

D) the pH at the equivalence point

Express your answer as a whole number.

E) the pH after adding 5.0 mL of base beyond the equivalence point

Express your answer using two decimal places

Any help with this is greatly appreciated! Thank you in advance!

Answer 1

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Both koh and Hbr are strong elechaly _ Initial ph. - pH = -log[n") = -10g [o-170m) — ph30:7707 2) from My, = M2V2 -_M - 0.170M W, = 35mi M-O-210M - Y2 = 0.1767 x 35m1 - 28-33 .1 0.2101 At equivaince point 28-3371 base sequired 3) mmol of HBr = 35XD-170 M, 5.95 mmol - b of kor = 101 m * 0-100 = 2,142 mmet HBr in excep =(5-95- 2.142) mol = 3.808mme [HT] - 3,803 menst - 0.0842M 45.2MI pke -lng [nt] - 0.0842] pH = 1:07) 1_gt equivalence point mmol of auld = nekat mmal obeys So as both are strong so it form a neutsal Solution having pH = 7.0 moment of KOH= 28.33 +Sm! 33.33 ml of love X10 mmol OEKAM - 32-330.210 6.99 mmol KOH is in excen (6.99-5-95mmal -04- Kon] -1.0mmolas +33.33 [OH) = 0.0152.M POHog Lon] - log O 0152 = 1-82 ich pH=14-pon = 14-1.82=/12.18]