7. In addition to its pharmaceutical uses, barbituric acid (HC4H3N2O3; abbreviated as HBarb; Ka = 9.8 x 10-5) has uses in textiles and plastics. Suppose a chemist was going to perform a titration analysis on 25.0 mL of 0.20 M HBarb using 0.18 M NaOH. This titration would require 27.8 mL to reach the equivalence point. How many moles of NaOH remains at the end of the reaction after the addition 10.00 mL?
a) 0.0 x 10-0
b) 1.8 x 10-3
c) 5.0 x 10-3
d) 3.2 x 10-3
Correct ans- Option d
In a titration, Equivalence point is that point where all the acid presentin the solution have completely been neutralized by the base added. On further addition of base after this point only make the solution basic as there is no more acid present for neutralization.
Now given the weak acid HC4H3N2O3 taken has Ka = 9.8 x 10-5
Now when we add a strong base like NaOH, then it react with the acid to from the salt of its conjugate base i.e
NaOH + HC4H3N2O3 --------------> NaC4H3N2O3 + H2O
It shows that 1 mole of NaOH is required to neutralize 1 mole of HC4H3N2O3
Now given HC4H3N2O3 taken = 25.0 mL of 0.20 M
That means mols of HC4H3N2O3 taken = concentration * volume
= 0.20 M * 25.0 mL
= 0.20 mol/1000 ml * 25.0 mL
= 0.005 mols
That means total mols of NaOH required to neutralize this = 0.005 mols
Now given titration would require 27.8 mL to reach the equivalence point.
That means in our 27.8 mL NaOH solution, we have 0.005 mols NaOH
Then in 10 ml of NaOH we have = 0.005 mols/ 27.8 mL NaOH * 10 ml of NaOH
= 0.00179 mols NaOH
So after addition of 10 ml addition, the rest of NaOH left = 0.005 mols - 0.00179 mols
= 0.00321 mols
= 3.2 * 10-3 mols
7. In addition to its pharmaceutical uses, barbituric acid (HC4H3N2O3; abbreviated as HBarb; Ka = 9.8...
A 35.0 mL solution of 0.0834 M barbituric acid (HA, pKa = 1.01, Ka=9.8 x 10-5) was titrated with 0.2500 M NaOH. Calculate the pH at the following volumes of added base Vb = 0.00mL Vb = 3.05mL Vb = 1/2 Vequivalence Vb = Vequivalence Vb = 12.5 mL
Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka = 4.0×10-10) by 0.429 M NaOH at the following points. (a) Before the addition of any NaOH _______ (b) After the addition of 5.30 mL of NaOH _________ (c) At the half-equivalence point (the titration midpoint) ________ (d) At the equivalence point ________ (e) After the addition of 30.7 mL of NaOH ________
Determine the pH during the titration of 73.1 mL of 0.462 M benzoic acid (Ka = 6.3×10-5) by 0.462 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 17.0 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 110 mL of NaOH
Determine the pH during the titration of 29.5 mL of 0.324 M acetic acid (Ka = 1.8×10-5) by 0.414 M NaOH at the following points. (a) Before the addition of any NaOH ________ (b) After the addition of 5.70 mL of NaOH _______ (c) At the half-equivalence point (the titration midpoint) _______ (d) At the equivalence point _______ (e) After the addition of 34.6 mL of NaOH ________
Determine the pH during the titration of 21.4 mL of 0.383 M benzoic acid (Ka = 6.3×10-5) by 0.338 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 5.90 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 36.4 mL of NaOH
A. Match each type of titration to its pH at the equivalence point. Weak acid, strong base Strong acid, strong base Weak base, strong acid pH less than 7 pH equal to 7 pH greater than 7 B. A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH. C. Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8 x 10^-5) with 0.20 M HNO3....
(1) Determine the pH during the titration of 58.4 mL of 0.386 M hypochlorous acid (Ka = 3.5×10-8) by 0.386 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 15.0 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 87.6 mL of NaOH (2) Determine the pH during the titration of 39.1 mL of 0.369 M ethylamine (C2H5NH2 ,...
Determine the pH during the titration of 20.1 mL of 0.383 M nitrous acid (Ka = 4.5×10-4) by 0.341 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 5.10 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 33.9 mL of NaOH
23) What is the pH of a solution prepared by mixing: 0.20 moles of acetic acid 0.40 moles of sodium acetate 0.10 moles of sodium hydroxide in 1.0 L of solution a) 4.74 b) 4.14 5.34 d) 13.00 e) None of the above 24) Barbituric acid (Ka = 9.8 X 10-5) is titrated with 0.200 M NaOH. What is the initial pH of 20.0 mL of 0.100 barbituric acida) 2.50 b) 4.01 c) 8.35 d) 7.00 e) None of the above 25) What is the pH in...
A student performs a titration of 25.0 mL of 0.100 M lactic acid (HC3H5O3), using 0.050 M sodium hydroxide (NaOH). The Ka for lactic acid is 1.4 x 10-4. a) (10 points) What is the pH of the solution after the addition of 23.5 mL of sodium hydroxide solution? b) (4 points) How many milliliters of NaOH are required to reach the equivalence point? c) (4 points) Write the chemical reaction that will determine the pH of the solution at...