Question

Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 2.58-M NaOH is added to 420. mL of each of the following solutions.

5. + 2/24 points Previous Answers My Notes Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 2.58-M NaOH is added to 420. mL of each of the following solutions. a) water pH before mixing = 7.00 pH after mixing= pH change = b) 0.132 M NH41+ pH before mixing = pH after mixing= pH change = c) 0.132 M NH3 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.132 M in each NH41+ and NH3 pH before mixing = pH after mixing= pH change =

Answer 1

Amount of water = 420.0mL. amount of 2.58 M Nach added = 10mL. final rowme of the sountion = (420+10) ML = 430mL final of Concentration – 10m2 x 2.58M 430mL = 0.06M. [OH-] log 0.06 - 1.22 010 pH = 14- polt = 14-1:22 = 12.78 pos pH before mining = 7.00 pH after mining pH change = = 12.78 5.78 dio Concentration of NH3t = 0.132M. As NHt is made up by the reaction of strong acid and weak base. so it got will be less than 7. Nit + H NHAT we know pkb of its slyl: 8x105) = -log 18 - logo = -0.25 +5 - 4.74 pH = 7 - 12 pro_ 1 loge = 7 - 4.74 – – 18g 8.132 = 7-2.37 to un = 5.07. When , Noh solution added, this will nutralize the Nhat solution. • Volume required for mitalisation - - 420.0ml x 0.132 M. - = 21.48 m2 2.58M

Therefore, when 10ml NaOH sow en the solution still be acidit. n 10m2 Nadh solution added & when some of 2.58M Naoth added, the solution Consist of unnutralised Holt and what. So, we have to calculate the concentration of salt [Nitat] and [Nit] 10ml x 2.58M [nits] = 10ml+420m2) m+420m2] = 0.06M. [Nitat] - 420m2 x 0.132M. = 0.129 M. 10m2 + 420m2 pH = 14-(pkb + log Csalt) = 14-(4.74 + log 063) - 1 .74 t 0:129 006 pH before mixing = 5:07 pt after mining = 8.90 - change = 3.83 a pH of the buffer solution is given by A oxbg I Ba [salt] s Base] = 14-(pkb + log 0.132). 0.132 = 14- 4.74 - 9.26

on when aoh added a then then the following reaction the following occurs NHAT +014 NaOH. so Nhat combines with OH-ion to form NHAGH. concontration of N hat decreases and NHz or Ni2017 increases. when, 10ml 2.58M Nath added, the no. of holes of OH- - 10 ml x 2.58 mol -0.0258 mal 1000 m2 0.0258 mol OH - Combines with 0.0258 may Nitta to form 0.0258 mot NHOCH. & Noot moles of Nhat before addition, 420m2 x 0.132 mol 1000 m2 -0.0554u mol. No. of moles of what after addition (0.05544 -0.0258) mos = 0.02964 mol. Limilarly, No. ots moles of NHGOH before addition. 420m2 x 0.132 mot = = 0.05594 nr. 1ooom No. of moles of Nita oh after addition = Co-055444 0.0258) nos = 0.08124 mol. 0.02964 I r volume in pH = 14- (Phot bg, 8.08 12 L same so direct use of moles gives - 9.69 pH before addition - 9.26 14 aftor addition = 9.69 same result 2