1. The K for FeO is 10* We mis two soutions of oqual volurne, cee containing 10

Question

1. The K for FeO is 10* We mis two soutions of oqual volurne, cee containing 10 M10, ons at 25"C. Which one of the follorwing a. A precipitate forms,because QK b. No precipitate forms because QK e. No precipitate forms becas Qk d. A precipitate forms, becese O K e. None of the other statements is true Chapter 19 Valees The following equilibrium constants will be useful for some of the problems Constant HNO, HOCI HF HCN H,So K-1.8x 104 K,-4.5 × 104 K,-3.5×10. K,-72x104 40 x 10 K,- very large HCO, (COOm CH COOH 42107 K-48 x 10 K-5.9x 10 KS-64 × 105 K-1.8×10-5 K,-3.5×104 HOCN -1.2x10 HOB K-2.5x 10 K-18 10s 12. Which of the following statements about titration curves is false? a. A titration curve for titrating a strong acid/strong base will always have an equivalence point at pH-7.0 The equivalence point of a titration is the point where chemically equivalent amounts of acid and base have reacted b. c. In titration curves for weak acid/strong base, pH changes near the equivalence point are too small for color changes to be used d. The equivalence point will be basic when titrating a weak acid with a strong base e. A titration curve shows how pH changes near the equivalence point. 13. Consider the titration curve describing the titration of a strong acid by addition of a weak base.Which following statements is false? a. At the equivalence point, the solution contains water and a salt. b. The pH does not increase very rapidly at the beginning of the titration c. The equivalence point is at pH

Explanation / Answer

1) the Ksp = 10-14

Now we will calculate Q and compare it with Ksp to know that whether ppt will form or not.

If Q>Ksp then ppt will form

If Q<Ksp no ppt will form

Q = [Fe3+][IO3-]3

Given:

[Fe3+] = 10-4M and [IO3-] = 10-5M [The concentration terms can be replaced with other conc terms to deterimine the formation of ppt , the image is not clear and it seems to be the same concentration as I used]

Q = 10-4 X (10-5)3 = 10-19

Q<Ksp, so no ppt will form.

2)

a)Titration curve of an acid and base will always have an equivalence point at pH =7, as they will form a salt which will not hydrolyze to change the pH of solution. [so the given statement is True]

b) The equivalence point of titration is the point where chemcially equivalence amount of acid and base have reacted: True

c)In titration curve of weak acid and strong base ,pH changes near equivalence point are too small for colour change to be used: False

d) true

e) A titration curve shows that how pH change at equivalence point: True

3) a) True, at equivalence point the solution will have water and salt of weak acid and strong base

b) True, as we are adding weak base to strong acid so pH will not vary at beginning

c) True, It will form salt of weak base and strong acid which will hydrolyze and pH will be less than 7.

d) False, pH will be lower at the beginning than at the end of experiment as we have started with strong acid

e) True

4) The solution will be a buffer of ammonia

the pH of buffer can be calcualted using Hendersen Hassalbalch's equation

pKb of ammonia = 4.74

pOH = pKb + log [salt] / [base]

pOH = 4.74 + log [NH4NO3]/[NH3]

pOH = 4.74 + log (0.45/0.45) = 4.74

pH = 14 - pOH = 14- 4.74 = 9.26

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