Question

Acid HX is a weak acid with Ka = 1.0 x 10–6

. A 50.0 mL sample of 1.00 M HX(aq) is titrated
with 1.00 M NaOH(aq). What is the pH of the solution at the points listed below during the
titration? For each question, write the letter of the correct choice from the choices given below.
A) 0.0 B) 1.0 C) 3.0 D) 6.0 E) 6.6
F) 7.0 G) 8.0 H) 9.85 I) 12.0 J) 13.0
12. Before any NaOH solution is added (i.e., what is the pH of 1.00M HX solution?) _________
13. After 25.0 mL of 1.00M NaOH has been added ....................................................... _________
14. After 40.0 mL of 1.00M NaOH has been added ...................................................... _________
15. After 50.0 mL of 1.00M NaOH has been added ....................................................... _________
16. After 51.0 mL of 1.00M NaOH has been added ....................................................... _________

Answer 1

pH formula's for salt's weak acid + strong baise salt tho pH at { pk w tp ka +Log e] strong acid + weak base salt tho pH = [ pk wapkb_loge]

HX + H2O → Hot + x ILI Ħolo -x 1 + x | +x El Tnxx [H, G+] [x] - [H*} 1x156 100 (1-x) = x² 106 - 100 (x) = x *²= 100x100=0 axe + bx tcro -bt]61-496 X xe - 166+ (2000)=461)(-106) x= Xam - 261) -10+0.02 0 .-106+0.02 - x= -106 = 0.0009995 X? [Hot] = 0.0009995 pH -10924, ot) = -log (0.000 2995) P = ( 4 )

(13) Hx: molarity=LM I Naot molarity= IM volume =50ml=0.052 volume=25ml ao.0252 Man I notot moles = 1X0-025 no-ot molesa 1x8.05 : = 0.025 20-05 - HXT Naot Naxt H₂O Ilo.05 0.025 0 - e-0.025 0.025 1+0.025 - élo.025 L o lan0251 Total volume = 005 + 0.02520.075L [4x] = > 0 - o.or5 [ Nax] = 0+025 = 0.3333 M - 0.075 k=l-0x1050 pha= -logla= -10g (1X188) 26 {Nax ] 0.025 0-3333 M pH = plkatlog & Hx] 0.3333 76 CD) 4-6 +109 özgaz

HXi Moloority = IM NaoH: Molarity = IM volumea soml=0.054 volumea 40mla 8.042 no-et moles = [X0-04 2004 not of moles i 1Xolos 20:05 fat Nadh > Waxt to I 0.05 0.04 To 1 -0.04 -0.04 1 tonda E 0.01 1 0 londy Total volume = 0.05+0.04 = 0.09L (Hx] = 0.0L goo 0.04 - zo.lill M . 0.4444 M Nax] = 0.09 PH , kệ + pH = 6 + log : 4029. 66 (E)

HX molarity LM | Nash molarity= IM volume = 50ml=0.052 volume = soml= 0.05L Ma 2 I noot molesalxooos no of moles = lxoos 2005 20-05 - HX + NaOH > Nay tho weak acid strong base salt Il 0.05 0.05 To 1-1 e-005 -0-85 +0.05 L o L 0-05 1 Total volume = 0.05+0.05-0.lk [Nax]> 005 2 0.5Ma [c] For this type of salt pH formula pH = + 1 pkt pka tloge pH- + 14+ 6 +log (0.5)) = pH = 9.8495 = 9.85 (H)

Ex: Molarity: IM NaOH molarity = IN I volume = 51mla 0.0512 volume=50ml = 0.052 no of moles = 1X0.05 20005 no of moles= 100.051 201051 Hx+ NaOH Nay + H₂O 10.05 |-0.05 El O loosi -0.05 0-001 Total volume = 0.05 toros) 20.2012 20.00.9901 M 0.00 [Nao#3= [04]+$. 3201 poh 709 2013-109 (0.002901) POH, 2.004 gha 14- pol a 14–2-004: 11.996 # = 1 ( I) - Rate please