Part C A 0.161 M weak acid solution has a pH of 4.29. FindKa for the acid. Expre
ID: 555893 • Letter: P
Question
Part C
A 0.161 M weak acid solution has a pH of 4.29. FindKa for the acid.
Express your answer using two significant figures.
The temperature for each solution is carried out at approximately 297 K where Kw=1.00×1014.
Part D
0.30 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution?
Express the pH numerically to two decimal places.
Part E
0.80 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 3.0 L of solution. What is the pH of this solution?
Express the pH numerically to two decimal places
Explanation / Answer
C)
we have below equation to be used:
pH = -log [H+]
4.29 = -log [H+]
log [H+] = -4.29
[H+] = 10^(-4.29)
[H+] = 5.129*10^-5 M
Lets write the dissociation equation of HA
HA -----> H+ + A-
0.161 0 0
0.161-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 5.129*10^-5*5.129*10^-5/(0.161-5.129*10^-5)
Ka = 1.634*10^-8
Answer: Ka = 1.6*10^-8
D)
Molar mass of HCl = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol
mass of HCl = 0.30 g
we have below equation to be used:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.3 g)/(36.458 g/mol)
= 8.229*10^-3 mol
volume , V = 4.0 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 8.229*10^-3/4
= 2.057*10^-3 M
So,
[H+]= 2.057*10^-3 M
we have below equation to be used:
pH = -log [H+]
= -log (2.057*10^-3)
= 2.69
Answer: 2.69
E)
Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol
mass of NaOH = 0.80 g
we have below equation to be used:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.8 g)/(39.998 g/mol)
= 2*10^-2 mol
volume , V = 3.0 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 2*10^-2/3
= 6.667*10^-3 M
So,
[OH-] = 6.667*10^-3 M
we have below equation to be used:
pOH = -log [OH-]
= -log (6.667*10^-3)
= 2.1761
we have below equation to be used:
PH = 14 - pOH
= 14 - 2.1761
= 11.82
Answer: 11.82
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