what is the pH of a solution after titrating 50.0 mL of 0.650 pyridine, C 5 H 5
ID: 895778 • Letter: W
Question
what is the pH of a solution after titrating 50.0 mL of 0.650 pyridine, C5H5N (Kb = 1.8x10-9), with 50.0 mL of HCl?
a. 4.47 b. 2.71 c. 5.74 d. 2.87 e. 5.43
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a 10.0 mL sample of 0.75 M CH3CH2CO2H (ka1=1.3x10-5) is titrated with 0.30 M NaOH. what is the pH of the solution after 22.0 mL of NaOH have been added to the acid?
a. 4.89 b. 5.00 c. 5.75 d. 4.12 e. 2.50
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what is the pH of a solution prepared with 0.50 M HC3H5O3 (lactic acid) (Ka1 = 1.4x10-4) and 0.75 M NaC3H5O3 (sodium lactate)?
a. 0.46 b. 2.00 c. 3.68 d. 4.03 e. 3.85
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calculate the solubility of sliver chromate , Ag2CrO4 (Ksp=1.1x10-12), in 0.005 M Na2CRO4.
a. 3.4x10-5 b. 1.0x10-6 c. 8.2x10-5 d. 5.2x10-7 e. 7.4x10-6
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If the solubility of barium fluoride is 7.2x10-3 M, what is the Ksp of BaF2?
a. 7.5x10-7 b. 3.7x10-7 c. 7.2x10-3 d. 1.5x10-6 e. 5.2x10-5
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which of the following aqueous mixtures would be a buffer system?
a. NH3(Kb=1.8x10-5), NaOH
b. H2SO4(Ka1=very large)(Ka2=1.2x10-2), CH3COOH(Ka1=1.8x10-5)
c. H3PO4(Ka1=7.5x10^-3)(Ka2=6.2s10^-8)(Ka3=4.8x10^-13), NaH2PO4
d. HCl, NaCl
e. HNO3,NaNO3
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How many grams of NaOBr should be added to 50.0 mL of 0.650 M of HOBr(Ka1=2.0x10-9) to make a buffer at pH 8.90
a. 123
b. 6.12
c. 8.70
d. 1.03
e. 1.58
Explanation / Answer
if total pyridine reacts with acid gives salt of weakbase,strong acid.
pH = 7-1/2(pkb+log C)
c = con of salt = 50/100*0.65 = 0.325
pkb = -log kb = -log(1.8*10^(-9)) = 8.74
pH = 7-1/2(8.74+log 3.25) 2.37
answer ; b.2.71
a.
No of moles of CH3CH2CO2H = 10/1000*0.75 = 0.0075 mole
No of moles of NaOH = 0.3*22/1000 = 0.0066 mole
so that the mixture will be buffer solution
pH = pka+ log(salt/acid)
pka = -log (1.3*10^(-5) = 4.88
pH= 4.88+log(0.0066/(0.0075-0.0066))
= 5.745
answer; c.5.75
b. it is a buffer mixture.
pH = pka+ log(salt/acid)
pka = -log ka = -log(1.4*10^(-4)) = 3.85
= 3.85+log(0.75/0.5)
= 4.026
answer; 4.03
Ksp= ((2S)^2)(S)
S = SOLUBILITY OF CrO4 = 0.005 m
Ksp= ((2S)^2)(0.005)
s = ((1.1*10^(-12))/4)^(1/2)
solubility = 5.24*10^-7 M
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