Question

The p?b values for the dibasic base B are p?b1=2.10 and p?b2=7.67. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.80 M B(aq) solution with 0.80 M HCl(aq).

What is the pH before addition of any HCl? pH=

What is the pH after addition of 25.0 mL HCl? pH=

What is the pH after addition of 50.0 mL HCl? pH=

What is the pH after addition of 75.0 mL HCl? pH=

What is the pH after addition of 100.0 mL HCl? pH=

Answer 1

*b = 7.948103 B(OH) ² BloH) + oH ههه kb =213%258 in auxion 0.8-2 - 0.8-150.8 as kb, is less !. 6,352x10-3 - 0163520102 a= 0.796 X10-1 Port = a= 0.0796 . POH = -log(0.0796) = 1099 PH= 147.099=12.901 @ 25ml Hl. moles (red) = 25 x08 -0.02 moles (B) = SO x 0.8 = 0.04 A B(OH)2 + 2 BloHt the froon moters 0104_ DOBS en X 1000 toot HU 0.09 0-02 Buffer is formed.

TpH = pky, & hogy 7 [B(OH) F = 2:10 + log on to - 2-10- 14-2.10 = 11.90 some Hel moles = 0.04 moles. equivalence point slott IHA - Blot)t + Hop 0.04 0.04 O 0 0.04 [Blou)] = x1000 = 0,4M pohte į pirka, tpikaz ) Tetor 267] = 4.885 - pH = 14-4.885 = 9.115 7 ul Hd mobes = 0.06 е (он), н* — Сон 21 4 И... - boy 0.06 and o.oy plo) + H BA? + Hoe 0.04 0.02 0.02 0.02 pOH = pka + log [872] a - [Bloht - 7.677 og 0,02 10-02 pH=14-767 = 7.67 6.33

classmate Date Page B(OH) + Hw - 0.04 B0H). +49 0.04 0.08 0.04 B/OH) ++ H+ oivu oon 0.04 Bt2 + How = a.266 ka, ako Bt² of Hao - Plon) + Ht 01966-52 ery) aty BOH) + Ho2 BIOH) + Ht (any) y (4+1) at na ang sa 2-13810R -0.266- A O-266 |_ 0.469 x 10-6- a? - 0.266 - a 20.1mm wiotc5 0.35*2x7023 Kay=y= lonth - 5.94x10- 537.8547072 Now [4+] = x= 3.52810-4 pH = 9-0.546 pH = 3.454.