Question

4. (16 Points) A) Calculate the pH of 0.250 Lof a 0.36 M Formic acid HCO2H and 0-30 M Sodium formate, NaCO, buffer. Assume that volume remains constant (K, for HCOH = 1.8 x 10^). B) Calculate the ph of the above buffer after the addition of a) 0.0050 mol of NaOH and b)0.0050 mol of HCI 5. (12 Points) a)Calculate the pH of 100.0 mL 0.20 M NH (K = 1.8 X 105). b)Calculate the pH of a solution made by mixing 100.0 mL 0.20 M NH, and 100.0 mL 0-30 M NH CI.

Answer 1

A 0.250L. given 0.36M formic acid PH 0.30M sodium formate. HCOH = 1.8x10-4 PH buffer. t : we know that by Henderson - Hasselblack eq?. pH = pka +log [ Base conjugate pH = -log[1:8x1024) + logf 0:30) - acid pH = 3.7447 + (-0·07904) - pH= 3.665 pH = 3.7 3.67. a) addition of 0.0050 mol. MaoH. - we need to determine moles of formic acid & Sodium formate after NaOH added. calculate amount befor add ng Maoh COOH - 10:36 ML) (0.250L -0.09 mol. HCOONa - (0.300 mily, (0.250 1) = 0.075 mol determine moles I MaoH. NaOH = > 0.0050 mole - (given Nach Eeact 1 molay Eatio with HCOOH

A COOH -> 0.09 mol-00050 mol = 0.084 mo COONa - 0 Osmo + 0005 Dol:= SOS DO. i ti Calculate pH. I j a pH = 3.67 tlog [ 0085 L. 0.08 --3-67+ 0 02632 PH= 3.696 PH = 3.74 +log 0.0 8 7 4 0.085 pH = 374 + (+62 -0.0263) - 3.7137 - pH = 3.71_ Scot b) after addition of 0:0050 mol HCL. HCOO. HE -> HCOOH. O moles a HCOO 3* = 0.25% 0:3= 0.005 - 0.07. moles of HCOOH 20-25X0:36 +0:00S - 0095 PH: pka tlog Ibase conjugate) [Acid pH = -log(1:8X154) + log (0.07/0.095) pH = 3.74+ (-0.1326) pH = 3.60 74. pH = 3.61

s calculate pH of tooo ml o oo m2 o.20M NH. (kb = 1.8x105) This is weak base solution. NH3(aq) +H2020) > MHz*(99) + OH (97). | ke 5 CHF *) (of). L3 XIA-S NH3 Initial change. Hits O x OH Alla 0. 20 -X 0.20-X. T equlibrium х (x)(x) = 1.8 X10-S. (0.20 -*) Assume X <0.20 x2 = 1.8 X1D-5 so X2= (0.20) (1.8 X10-5) 0:20. = 3-6*10=6. - 36 X 106. 1-23 X = (3.6 x 10-62. - 1.9x10-3 pol= -logo (1.9810-3) = -272 pH = 14.00- 2.720. 1:28.

DUI Page 80- b) MH4Cl is soluble ionic compound 3 when added to water undergoes - NH4Cl (5 ) HH4 (1) + c/(1). initially contains 0.20th NH 3 4 0.30M NH4+. This is buffer solo : Henderson.e99. base pH = pka + log to tteid - ka for NH4 = 1.0X10-14 - 10 x 10-14 1.8x 10-5 pH = -logelo) (5.6x10-10) + logetot (0.2 0.30 = 9.25+ (01760) - 9.074 kb 5.6X1o -LO.