utak acid 8. (15 points) A solution of 30.00 mL of 0.1901 M HA (aq) is titrated
ID: 551716 • Letter: U
Question
utak acid 8. (15 points) A solution of 30.00 mL of 0.1901 M HA (aq) is titrated with 0.2309 M NaOH (aq). The constant K, for HA is 42 x 10 KalátetheintilsHof]|IADsolution.sth)(x) ·(braol )-HaxlpW-X2 PH- log (o.0028 256 322) - 2.S pH-aS p H B. What volume of NaOH (aq) will be required to reach the equivalence point? ouc 1000mL ,2309 0.1a01 HA- mole s HA = .005203 noles #A -- 2304 0.03L 40.48 mL or .040S L Nao will be requrred . What is the pH of the solution at the equivalence point? naycal mo lanty of A nols RA wiiailu pcsent Liters Of Solnt10 (asi.) + (.OZ.47 ) ? Nath Kw ('19 HA ) ( .Explanation / Answer
C) The reaction between HA and NaOH is HA+NaOH--> NaA + H2O
At the equivalence point the concentration of A- = (0.1901)(0.03)/(0.03+0.0247)
= 0.10425M
Ka of solution = 4.2x10^-5 so pKa = -log Ka= 4.376
pKb= 14-pKa = 14-4.376= 9.624
Kb= 10^-9.6 = 2.5x10^-10
[OH-]=square root of (KbxC)= (2.5x10^-10x0.104)^1/2=0.5x10^-5
pOH= -log (0.5x10^-5) = 4.7
so pH = 14-pOH= 14-4.7= 9.3
d) after adding 8.5mL NaOH
moles of NaOH in 8.5 ml = (0.0085 x 0.2309)=1.96x10^-3
moles of OH- initially present in 0.0547 L = 0.5x10^-5
Total moles of OH- = 1.96 x10^-3 approximately ( other one negligible)
Total volume = 0.0547+0.0085= 0.0632
so [OH-]= 1.96x10^-3/0.0632 = 0.0310= 3.10x10^-2
pOH= -log [OH]= -log3.10x10^-2 = 1.51: pH=14-pOH= 14-1.51 = 12.49
e) After adding 37.20ml NaOH
Number of moles = 0.0372 x 0.2309 = 8.6x10^-3
Total volume = 0.0547+0.0372= 0.0919L
so [OH]= 8.6x10^-3/0.0919 = 0.094
pOH= -log[OH]= -log 0.094= 1.02 and pH= 14-1.02 = 12.98
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