Strong base added during the titration of 10.0 mL of an unknown acid. The graph

Question

Strong base added during the titration of 10.0 mL of an unknown acid. The graph shows the acid is weak and triprotic. weak and diprotic. strong and triprotic. strong and monoprotic. Which of acetic acid solution using sodium hydroxide solution? An aqueous solution of the salt NaA is added. What happens to the pH of the solution? It decrease because additional acid has been added. It stays the same because the NaA solution is neutral. It increase because the conjugate base has been added. Calculate the pH of an aqueous 0.15 M benzoic acid solution. 2.5 4.2 5.0 6.3^1_1H +^2_1H rightarrow^2_2He +^1_0 pi^2_1H +^3_2H rightarrow^4_2He +^1_1 pi^1_2H +^1_3H rightarrow^2_4He +^0_1 pi^2_1H +^3_1H rightarrow^4_2He +^1_0 pi The half-life of radium in a solution of radium nitrate is longest in 100 mL of a 1.0 M solution longest in 1000 mL of a 1.0 M solution longest in 100 mL of a 0.5 M solution. the same in each of the solution in (A), (B) and (C). This graph represents the decay of a radioactive isotope. What is its half-life? 2hr 3hr 4hr 6hr Which reaction does not form a new element?^23_10 Ne rightarrow ? +^0_-1e^2_1 H +^2_1H rightarrow ? +^1_-0 pi^140_50 Ce +^1_0 n rightarrow ? +^0_0 gamma^10_5 B +^1_0 n rightarrow ? +^4_2 He

Explanation / Answer

Answer – 34 ) We are given the acetic acid and sodium hydroxide solution. We know acetic acid is weak acid and sodium hydroxide is strong base, so this titration is in between weak acid and strong base. So as we added NaOH ther is pH increase slowly and there is formed first half equivalence point where pH = pKa and then there is sudden change in pH and again it gets increase slow and then gets stable. So there is option D is correct titration curve.

35) We are given the weak acid HA with pH is 4.0 and when we added salt NaA then there is conjugate base and acid reaction and salt converted to acid, means there is conjugate base added, so pH gets increased, but there is formed buffer solution of HA and NaA, so there is D) It increase slightly because the acid has been diluted

36) We are given, [C6H5COOH] = 0.15 M , Ka = 6.1*10-5

We need to put ICE chart

C6H5COOH + H2O ------> H3O+ + C6H5COO-

I 0.15                                   0             0

C -x                                   +x           +x

E 0.15-x                            +x           +x

Ka = [H3O+] [C6H5COO-] / [C6H5COOH]

6.1*10-5 = x*x /(0.15-x)

We know the Ka values of benzoic acid too small, so we can neglect x in the 0.15-x

6.1*10-5 *0.15= x2

x = 3.02*10-3 M

so, x = [H3O+] = 3.02*10-3 M

so, pH = -log [H3O+]

           = -log 3.02*10-3 M

           = 2.52

So answer is – A) 2.5

37) We are given the diagram nuclear reaction and in that proton and neutron show by shaded ball and empty ball.

There is first 1 proton and 1 neutron reacts with 2 neutron and 1 proton means reactant has 21H and 31H and it gives 2 neutron and 2 proton molecule and 1 neutron, so product side there is 42He and 10n, so balanced reaction as follow –

B) 21H + 31H ----> 42He + 10n

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