Question

You have 1.25 L of a 2.00 M solution of Hypochlorous acid, HClO to which you will add 1.5 M NaOH to adjust the pH.

a) what is the pH of hte solution when 0.500 L of NaOH is added?

b) How many mL of soldium hydroxide should be added to prepare the 7.00 pH buffer solution?

c) If 75.0 mL of 1.5 M Ba(OH)2 is added to the buffer in part 'b' when it is starting at pH 7.00 before adding Ba(OH)2, what is the new pH? Ba(OH)2 is a strong dihydroxic base.

Answer 1

1.25 L of 2.oom hyper chlorous acid = 1.25XX 2.00 ml of hyperchlorousund. = 2.5 mol of hyperchlorous aced. @ 0.57 ofis m nuo i = 0.54x 1.5m? A wwolt = 0.75 mol of woolt. 0.75 mol of mooh react with 0.77 ond of and to form 0.77 mol of salt. therefore the soln contain (2.5-0.75) 19 1.75 mol of exun acid tsalt) = 0.7506 (1:25+0. 5 1 0.75 m 1.75 taud] = 1.75 mol (lastot i = hos m. from the literature it is observed that pka value for hyperchlorom and is 7.53. than, from thenderson ear, pH = pka + wg taud taut)

0.75/1.75 thus from eae we have pH = 7.53 + log - 1711-7- = 7.53 +log at 57-16 () ut, v rool of Nach should be added. me of 1.8 m noolt V kx 1.5 mol of woole - 1.50 mol of Naoll. lisrmol of Naoh react with 1.5v mol of and to form I.TV mol of salt. The refore the sin contum (2.5-1.50) mol of exum aced. (1.250 L taud] =(2.5-1.56 mol (2.5-1.50 (1.25 + V M m [salt] = 1.55 mol (1.25tv) L (1.25 + -Now from ear we nang 7=7+5+ wy- 1.5V 1-25 2.t-litr (lyiy - 2.5-1.5V 7-7-53 = log 0 a Log 15 gr = -0.5 urv = 0.30 133, 15V = 0.75 – 0-45V 3 (1.5 +0.45) v = 0.75 or= 0.3852 = 385 m2. 385m 1 of molt should be added

(C) Bacon) ----> Balt+- 2014 - 1.5m 2x1.5m Bacom - -1.5m : LOH) = 2x1m = 3m - 0.385x1.5ml = 0.5775 mok. from the port (6) til obroved that the sola contam (2.5-1.5x 0.385) mol of , 1.9225 mol of exam alld. [Aud) = 1.9226 mol And rootmins of (1.25+0.385) --- T-18 M 75m1 of 3 m oth = 752 x 320 tour - 0.225 mol of our Urmol of of react with our mol of sound oh-to form 0.225 ml of salt. thin the schn contra (1.9145-0.227) I 1.6775 mol of exem celd. therefore toful salt = co.nr + 0.5775) mol = 0.8o5 mil acone of salt = 0.8025 mol e opow otal volume of Hem 0.8o2rmof V = _0.8027 conc. of Had _19224-7 1-9227 M L whom w=1larto.385 + cort = 1.7L This from eak o we have, 0.08028/ pH = 7.53+ wg 1 .9215/61 = 7.53 tlg 0.0025 1-9225 = 7.15