Please help with these 7 questions....From 7 to 13 please. 7. What is the pH of
ID: 714001 • Letter: P
Question
Please help with these 7 questions....From 7 to 13 please.
7. What is the pH of a solution with a. H' 0.1M b. H' 2.5 x 10 "M c. H 4.6 x 10°M 8. What is the pH of a solution with a. OH 01M b. OH = 5.6 x 103M c. OH 3.8 x 109M What is the concentration of an HCI solution if 35mL of HCI are neutralized by 45mL of 0.1M NaOH? 9. 10. What is the concentration of H,SO, if 82mL of 0.2M KOH is needed to neutralize 20mL of H,5O4? 11. What is the molarity of a solution made by dissolving 3.4g Ba(OH)2 in enough water to make 450mL of solution? with 0.5 moles of HC2H,O2 and 0.5 mol Na CH302 in enough 12. What is the pH of a buffer made water to make 1L of solution? 13. What is the pH of a buffer made with 0.65M formic acid and 0.55M formate?Explanation / Answer
7)
a) [H+] = 0.1M
-log[H+] = -log(0.1)
PH= 1.0
b)
[H+] = 2.5x10^-4M
-log[H+] = -log(2.5x10^-4)
PH= 3.60
c) [H+] = 4.6x10^-8M
-log[H+] = -log(4.6x10^-8)
PH= 7.33
8)
a) [OH-] = 0.01M
-log[OH-] = -log( 0.01)
POH= 2.0
PH+POH= 14
PH= 14-POH
PH= 14-2.0=12
PH= 12
b) [OH-] = 5.6x10^-3M
-log[OH-] = -log(5.6x10^-3)
POH= 2.25
PH= 14-2.25=11.75
PH= 11.75
c) [OH-] = 3.8x10^-9
-log[OH-] = log( 3.8x10^-9)
POH= 8.42
PH= 14-8.42
PH=5.58
9)
NaOH + HCl ------------------ NaCl + H2O
1 mole 1mole
NaOH HCl
Molarity=M1=0.1M Molarity = M2=
Volume=V1=45mL Volume= V2= 35mL
number of moles=n1=1 mole number of moles=n2= 1 mole
for neutralisation
M1V1/n1= M2V2/n2
0.1x45/1= M2x35/1
M2= 0.128M
Concentration of HCl = 0.128M
10)
2 KOH + H2SO4 ---------------- K2SO4 + 2 H2O
2 mole 1 mole
KOH H2SO4
Molarity = M1= 0.2M Molarity = M2=
Volume = V1= 82 mL Volume= V2= 20 mL
moles = n1= 2 mole moles= n2= 1 mole
M1V1/n1= M2V2/n2
0.2x82/2= M2x20/1
M2= 0.41M
Concentration of H2SO4 = 0.41M
11) mass of Ba(OH)2 = 3.4 grams
Volume= 450 mL = 0.450 L
molar mass of Ba(OH)2 = 171.34 gram/mole
number of moles = mass/molar mass = 3.4/171.34=0.0198 moles
Molarity = number of moles/volume in L
Molarity = 0.0198/0.450 = 0.044 M
Molarity = 0.044M
12)
number of moles of HC2H3O2= 0.1 mole
number of moles of NaC2H3O2 = 0.1 moles
PKa of HC2H3O2= 4.76
PH= PKa+log[salt]/[acid]
PH= 4.76+log(0.5/0.5)
PH= 4.76
13)
Concentration of Formic acid= 0.65M
Concentration of formate= 0.55M
PKa of formic acid= 3.8
PH= 3.8+log(0.55/0.65)
PH= 3.72.