Question

Write legibly. Show your work for full credit. All calculations should be performed for an aqueous solution at a temperature of 25 °C. 1. Calculate the pH after the addition of 0.00, 10.00, 25.00, 40.00, 45.00 ,50.00, 51.00, 55.00, 60.00, and 70.00 mL of 0.1000 M of HCl in the titration of 50.00 mL of 0.1000 M NaOH.. 2. Make a graph of pH versus the volume of HCl added. (Use Excel Spread Sheet)

Answer 1

pH 0 10 20 60 70 80 30 40 50 Vol. of HCl added(ml)

1. at 25°C We are tétrating 50 ml of 0.1000M NaOH by 0.1000 of Ha. 1 addition of some of 0.1000M of Ha: I [H0] = 0.1000 M pot = -log Loto] 8. PH a 14-POH = 14 -1 = 13 (ii) addition of 10.00 me of 0.14 of Heli moles of Ho - (0.1 mokſlit) x 10.00 ure = 1 mole x me 100 moles, lit 50 ure of 0.1 M NaOH contains = (50 80 1x108) moles. 58103 moles equivalent no. of 40 neutralizes equivalent no. of oHo cons. & excess oHo iows = (5x103_163) moles 2 4x108 moles. in concentration of excess oHo ions = 4x100 moles 20.0667 male (50+10) me & pot = log [o Ho] - log (0.0667) & pH a 14 - pH = 14+ log (0.0667) = 12182 (iii) addition of 25 ml of 0.1 M of Heli moles of Ho ious = (25 mLX0-1 moles (lit) = 2.5x100 moles excess of CH® ions : ( 5x160 - 2.5 X163) = 2.5 17moles excess concentration of the ions - 2.5 x 103 moles are 5. 34- pH = 54+ ley Cotº) = 54 + log (00335) = 12:52 Н (50+25) me = 010333 M

0.0111 H (50+40) (iv) addition of 40 ml of 0.1M of hai moles of Ho ious - (40mLy on moles/ lit) & 4x108 moles excess of oHo ions = (5x10 3 _ 4x103) ruotes = 10 moles concentration of excess oHo ious = 100 moles :. PH = 14-poll : 14+ log Cohe) = 14 +log (0-011) = 12:04M (v) addition of 45 ml of 0.1M of Holo moles of H® fous = ( 45 ml X01 molesllit) = 45X16% moles excess of our cousa (5x1024.5x103) 2 015 8103 moles concentration of excess oH® ions : 0:5X16 Prioles5.263x17% M (50+45) me HopH = 14-poh = 54+ log [04°) - 14 +log (5-263x16) = 11-12 (vi) addition of some of olM of Hl: moles of He cows = (50 ml x 0:1 notes / lit) = 5x108 moles no. of moles of Ho cons= no. of moles of oHo ions. Hence no excess Ho or OH ions. pH=7 (vii) addition of 50 ml of 0:1 M of Hu: watch moles of Holous = (51ml. XDil molesllit) - 5:1X16 moles excess of H® ions = (5.4x1025 x 10 ) = 0.1X16% moles. concentration of Melons = Oil xt6 moles (50 +51) me = 9.90099x107 2. pH =- 199HⓇ] =-bog (9.90099x10-2) = 6.00 (vii) addition of 55 mb of 1 M of HUL: moles of HⓇ ious - (5524X01 moles Idit) - 5.5x173 moles excess of Ho cons= ( 5.5x10 35x103) 2 0.5 x 103 moles concentration of Holons 2 0.5x108 mokes] = 4.7619x108m ...pH = -- og CHⓇ 1 - - 10g [4-7619 36 37 = 2:32

(ix) addition of 60 ml of all of the moles of Ho cons = (60 ml x oil notes/lt) = 6*103 moles excess of H® cous = (6x108 5x103) moles = 16 moles Concentration of H® cons 108 moles = 9.0909 8 X 15 3M 0.0167 M (60+50) me ..pH = -log [HⓇ] = -log (9:0909 X10%) = 2:04 (x) addition of toue of 0.1 M of Helo moles moles of Hoions (70 nexo 1 moles (lit) = 7x10 excess of to cous = (4x108 5x103) moles = 2x10 moles concentration of Ho cons a 2x103 moles non (70+50) ut pH = -log [H0] = -log (0.0167) = 1178 2. Table for graph plotting I yolume of HCl added, I PH 13.00 0.00 10.00 12.82 25.00 12.52 40.00 12.04 11.72 45.00 50.00 7.00 51.00 6.00 55.00 2.32 60.00 2.04 7000 1.78