1. Would the bromocresol green indicator with a transition range of 3.8 to 5.4 e
ID: 700819 • Letter: 1
Question
1. Would the bromocresol green indicator with a transition range of 3.8 to 5.4 ever be of use when titrating a weak acid with a strong base? Explain. 2. Explain the reason a blank needs to be performed in most analyses 3. Outline the technique used to prepare a buffer. Note that I will often give specific salts and acids/bases and that if I do you can calculate out specific amounts to mix 4. Two buffers are prepared to the same volume. In the fist, 0.25 mL of concentrated weak base, and 0.30 grams of the salt of the weak base were used. In the second, 0.50 mL of concentrated weak base and 0.60 grams of the salt of the weak base were used. Compare the pH and the buffering capacity of the two buffers prepared 5. Explain the problem(s) with the statement "an acid can be added to a buffer without changing the pH until the buffer is exhausted". 6. Explain the reason buffers are important in chemistry Problem 1. Determine the amount of 0.1125 M NaOH needed to completely neutralize 20.00 mL of 0.03561 M phosphoric acid. Find the pH at the equivalence point ANSWER: 18.99 mL: 12.7ish Titration of 0.824 grams of KHP (204.22 g/mol) required 38.314 grams of NaOH solution to reach a 2. phenolphthalein endpoint. Find the weight molarity of the NaOH solution. ANSWER 0.105 mol NaOH/kg solution gram aliquot of H2SO4 solution required 57.911 grams of the NaOH solution to reach the phenolphthalein 3. Based upon the solution in problem number 2, find the concentration of H,SO, solution in mol/kg if a 10.063 endpoint. ANSWER 0.302 mol H,SOJ kg solution Use the Henderson-Hasselbach equation to fing the pH of a buffer made from 25.00 mL of 0.1000M acetic 4. acid and 0.35 grams of sodium acetate. ANSWER 4.99Explanation / Answer
Problem 1
Balanced chemical reaction
H3PO4 + 3 NaOH = Na3PO4 + 3 H2O
Moles of H3PO4 = molarity x volume
= 0.03561 mol/L x 0.020 L = 7.122 x 10^-4 mol
Moles of NaOH required = 3 x Moles of H3PO4
= 3 x 7.122 x 10^-4 mol = 2.136 x 10^-3 mol
Volume of NaOH
= (2.136 x 10^-3 mol x 1000 mL/L)/ (0.1125 mol/L)
= 18.99 mL or 19 mL
Volume of solution = 20 + 19 = 39 mL
Moles of Na3PO4 formed = Moles of H3PO4 consumed
= 7.122 x 10^-4 mol
Concentration of Na3PO4 = 7.122 x 10^-4 mol/ 0.039L
= 0.01826 mol/L
Reaction of formation of OH- ions
PO4 3- + H2O = HPO4 2- + OH-
I 0.01826 - - -
C - x - +x +x
E 0.01826-x - x x
Equilibrium constant expression for H3PO4
Ka = (10^-14) * (0.01826-x) / ( x * x)
2.2 * 10^-13 = (10^-14) * (0.01826-x) / ( x^2)
x = 0.071 M = [OH-]
pOH = - log [OH-] = - log (0.071) = 1.15
pH = 14 - pOH = 14 - 1.15 = 12.85
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