c. The plH of the the 25.00 ml. with 28.00 ml. of the methylamine solution. mL o
ID: 569971 • Letter: C
Question
c. The plH of the the 25.00 ml. with 28.00 ml. of the methylamine solution. mL of the methylamine solution d. The pll of the solution after 25.00 ml. of the h Department .6 Chemistry and Biochemistry HAM 116 Spring 2018-QIZ 4 part1 hat & da: Phan read the gusefully Port 1 it wertka otal of 40 poits NPoints nt of 100 pointe otal). Phae wvite your rome on the back. Good Luck HCI 1. In a titration experiment a student titrated 25.00 mL of 0.1120 M hydrochloric acid (HCI) with a 0.09981 M calcium hydroxide (Ca(OH)2) solution. Please calculate the following: a. The pH of the hydrochloric acid solution. b. The pH of the calcium hydroxide solution. c. The pH at the equivalence point. d. The pH of the solution when 4.00 mL of the calcium hydroxide solution were added to the 25.00 mL solution. The pH of the solution after 25.00 mL of the hydrochloric acid solution were mixed e. with 16.00 mL of the calcium hydroxide solution. f How many mL of the calcium hydroxide solution are needed to neutralize the 25.00 mL of HCI Write the neutralization reaction. g. . Plot the pH titration curve (see example on SAKAI under Resources). 2 2, 2 HBNHExplanation / Answer
(a)pH of 25 ml .1120M HCL is 2.55
calculation;
1000ml of HCL solution contain [H+] moles =.1120
1 ml of HCL solution contain [H+] moles=.1120/1000
25 ml of HCL solution contain [H+] moles =(.1120/1000) *25
=2.8*10^-3
now pH =-log[H+]
=-log[2.8*10^-3]
=-[log2.8+log10^-3] ;(log a*b=loga+logb)
=-[0.45-3]
=2.55 answer
(b)pH of 0.9981 Ca(OH)2 will be =13.3
calculation ;
ionisaton of one mole Ca(OH)2 gives one mole of Ca2+ions and 2 moles of (OH-) ions
then 0.09981 mole of Ca(OH)2 will give =0.09981*2=0.19962 moles of (OH-) ions
it is base so first calculate pOH
pOH=-log[OH-]
=-log[0.19962]
=-{log19962/10000}
=-{log19962+log10^-5}
=-{4.3-5}
=0.7
now pH+pOH =14
pH = 14 -pOH
pH =14-0-7
pH=13.3answer
(c)volume of calcium hydroxide is not given.
(d)Normality = Molality*n
in case of acid n = basicity
for HCL n=1
N= M=0.1120
milli-equivalents = 0.1120*25=2.8
for calcium hydroxide N= M*n ( n= acidity)
N=0.09981*2=0.19962
milli-equivalents of calcium hydroxide=0.19962*4=0.8 approx.
no of milli-equivalents of acid left after the addition of base=2.8-0.8=2
total volume of solution = 29ml
; 29ml of solution contain =2 milli-equivalents of HCL=2*10^-3 equivalents
1 ml of solution contain = 2*10^-3/29
1000 ml of solution contain= (2*10^-3/29)*1000 =0.06equivalent of HCL
0,06N HCL=0-06M HCL
pH=-log[H+]
=-log[0,06]
=1.3 answer
(e) In HCL normality = molarity= 0.1120
volume of acid used =25ml
milli-equivalents of acid = 0.1120*25=2.8
; For calcium hydroxide normality =2*molarity = 2*0.09981= 0.19962
volume of base used = 16ml
milli-equivalents of base =0.19962*16=3.19 approx
no of milli-equivalents of base left when it is added into acid = 3.19-2.8=0.39
;total volume =16+25=41ml
41 ml of solution contain =0.39 miliequivalents of base=0.39*10^-3 equivalents
1 ml of solution contain=0.39*10^-3/41
1000 ml of solution contain=(0.39*10^-3/41)*1000=7.3*10^-3
molarity =normality /2(in case of calcium hydroxide)=7.3*10^-3/2=3.75*10^-3
pOH =-log[OH-]
=-log[3.75*10^-3]
=2.5
pH=14-2-5
=11.5answer
((f) 14 ml of Ca(0H)2 required for complete neutralisation
(g) Ca(OH)2 + 2HCL -> CaCl2 + 2H2O
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