ilr nyuloxiue solution before it is 4. What purpose does the indicator, phenolph
ID: 541409 • Letter: I
Question
ilr nyuloxiue solution before it is 4. What purpose does the indicator, phenolphthalein, serve in the titration? 5. What are equivalence points and end points and how do they differ? 6. What is the molarity of a solution that contains 2.47 g of H2SO, in 0.350 L of solution? 7. Complete and balance the following equations: a) Ba(OH)2(s)+HNO3(aq) b) KOH(aq)+H3PO4(aq)-> 8. a) How many milliliters of 0.20M HNO3 are needed to completely neutralize 35.0 mL of 0.10M Ba(OH)2 solution? b) How many milliliters of 3.0M H3PO4 are required to neutralize 75.0 gof KOH?Explanation / Answer
Phenolphthalein is an idicator that undergoes a colour change from colourless to pink that begins at PH of 8. in this titration your performing the phenolphthalein will start to change colour at the point when the moles of acid equal the moles of base. Although this colour change occurs at PH of 8 and not at PH of 7.The phenolphthalein is commonly used because of the distinctive colour change that occurs. Phenolphthalein is colourless in acids and pink in base.
5.The equivalence point is where the amount of acid equalt ot amount of base in the solution. Thus the solution is neutral PH = 7. At this point no indicator work.At equivalence point PH reamins the constant. The very small amount of more acid/base is added change in pH . The change PH is called end point of the titration.
6. molarity = W/G.M.Wt * volum of solution in ml
= 2.47/98*0.35 = 0.072M
7. a. Ba(OH)2(aq) + 2HNO3(aq) ---------> Ba(NO3)2(aq) + H2O (l)
b. 3KOH (aq) + H3PO4(aq) -----------> K3PO4(aq) + 3H2O(l)
8. . Ba(OH)2(aq) + 2HNO3(aq) ---------> Ba(NO3)2(aq) + H2O (l)
1 mole 2moles
Ba(OH)2 HNO3
M1 = 0.1M M2 = 0.2M
V1 = 35ml V2 =
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 0.1*35*2/0.2*1 = 35ml
b. 3KOH (aq) + H3PO4(aq) -----------> K3PO4(aq) + 3H2O(l)
3 moles 1moles
no of moles of KOH = W/G.M.Wt
= 75/56 = 1.34 moles
3 moles of KOH react with 1 mole of H3PO4
1.34 moles of KOH react with = 1*1.34/3 = 0.45 moles of H3Po4
no of moles of H3Po4 = molarity * volume in L
0.45 = 3* volume in L
volume in L = 0.45/3 = 0.15 L = 150 ml >>>>answer
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