Question

Part A A beaker with 165 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 mol L- A student adds 6.70 mL of a 0.320 mol L'HCl solution to the beaker. How much will the pH change? The pK, of acetic acid is 4.780. Express your answer numerically to two decimal places. Use a minus ( - ) sign if the pH has decreased. View Available Hint(s) EVO ADO ? ApH- Submit

Answer 1

ANSWER

ANSWER From the question given, we have m) Total Volome pH 8 the o acetr and acon acid v= 165 a 5.00 Total Molarity of aud & ConTugabe base = 0.) moll. = M. Molaruby & And He added = 0.32 mork' - Muel Volume & Hol added & b.tml - Vuo pea & alltr and a 4.780 - pra for the total solarul if x P lhe molarity acid, this molaruby & Consugate base -- x = 0.1-x. Now From Henderson - Hase / batch equation : (equation to determine ratio & Concentration of weak alid and its conregate base in butter solution Maw popka + lay (base) ., 5. 4.780 + log (open) 78 0. x7 o 5.4.780 = 0.22 2

0.1 - 2 = 1.66 x 2.66 x = 0.1 or x = 01/2.66 Molaru by Molarsby aled & base & 0.037 M. = (-0.037) - 0.063 in Mo larul is 10 & and 1000 w) 165 165m) is a 0.037 s. 0.037 x 165 Ma = 6.105x 103 m Molaruby is & base is lovo ) 10 165 m 65 ml - : 0.063 s. 0.063 & 165 m lovo 10.395 x 103 m M = Now when the student adds 6.7 / 9 0.32 M HCl, Molarity of Hel is [Mha= 6.7 s is, 0.32 x 67 m - 21/tydex

Hence di net molarity of and a [u]. Ma-Mael - 6:105 x 103 - 20/94x100 m) (wit) – 3.961 x 10² si molardly of acid, After the addition, the new Et MH. 3-967 x 63m Now the pt after adding reel as pro ph' - – log (3-961 x 103 m] Hence change in apH-5-2.9. se ph decreases by 2.6 ANS

The calculation is made fron the Henderson-Haselbalch equation used for the determination of ratio of weak acid and its conjugate base in buffer solution and basic molarity concepts.