PCAT - General Chemistry Problems [XI-XII]

The half-life of radioactive sodium is 15.0 hours. How many hours would it take for a 64 g sample to decay to one-eighth of its original activity? A. 3. B. 15. C. 30. D. 45.

D. For 64 g of a sample to decay to 1/8 of its original activity, or 8 g, the sample would have to go through 3 half-lives. Therefore, the amount of time needed for the decay is 3 half-lives x 15 hours per half-life = 45 hours.

Which of the following combinations would produce a buffer solution of pH = 4? (Ka HNO2 = 4.5 x 10^-4). A. 0.30 M HNO2, 0.22 M NaNO2. B. 0.22 M HNO2, 0.30 M NaNO2. C. 0.11 M HNO2, 0.50 M NaNO2. D. 0.50 M HNO2, 0.11 M NaNO2.

D. See the attached image for the work.

What is the ratio of [H+] of a solution of pH = 4 to the [H+] of a solution of pH = 7?

→ Subtract pH: 7-4 = 3. → This means 10^3. → Or 1,000:1.

In an exponential decay, if the natural logarithm of the ratio of intact nuclei (N) at a time t to the intact nuclei at a time t=0 (N°) is plotted against time, what does the slope of the graph correspond to?

-λ. The expression N= (N°)e^-λt is equivalent to (N/N°) = e^-λt. Taking the natural logarithm of both sides of the latter expression you find: ln(N/N°) versus t will give a straight line of slope -λ.

A patient undergoing treatment for thyroid cancer receives a dose of radioactive iodine 131-I, which has a half-life of 8.05 days. If the original dose contained 12 mg of 131-I, what mass of 131-I remains after 16.1 days?

3 mg. Given that the half-life = 8.05 days, we know that 2 half-lives have elapsed after 16.1 days, meaning that 25% of the original amount is still present. Thus, only 25% of the original number of nuclei remain. Since the dose is 12 mg, only 3 mg will remain after 16.1 days.

Which of the following sets of materials would make the best buffer solution? A. H2O, 1 M NaOH, 1 M H2SO4. B. H2O, 1 M HC2H3O2, 1 M NaC2H3O2. C. H2O, 1 M HC2H3O2, 6 M NaC2H3O2. D. H2O, 1 M HC2H3O2, 1 M NaOH.

A buffer solution is prepared from a weak acid & its C.B., preferably in near-equal quantities. Choices A & D do not show conjugate acid/base pairs, so they are incorrect. Choice C is incorrect because the deviation in concentration, meaning B is the correct answer.

For each of the following pairs, choose that which describes the weaker acid: A. Ka = x ; Ka = 3x. B. [H+] = x ; [H+] = 3x. C. pKa= x ; pKa = 3x. D. pH = x ; pH = 3x.

A. Ka = x. → Ka is a measure of dissociation of an acid, & therefore the strength. A higher Ka = stronger acid. x is one-third the value of 3x, & therefore, a weaker acid. B. [H+] = x. → [H+] is a direct measure of the strength of an acid. The greater the concentration, the stronger the acid. x is smaller than 3x, making that the logical answer. C. pKa = 3x. → pKa = -log[Ka]; therefore, a pKa of 3x corresponds to a Ka lower than that of a pKa of x. A lower Ka means a weaker acid, & a lower pKa means a stronger acid. D. pH = 3x. → The acid with a pH of 3x is the weaker acid, using the same reasoning as above.

At equilibrium, a certain acid, HA, in solution yields [HA] = 0.94 M & [A-] = 0.060 M. A. Calculate Ka. B. Is this acid stronger or weaker than sulfurous acid (Ka= 1.7 x 10^-2)? C. Calculate Kb. D. Calculate pH. E. Calculate pKa.

A. Ka is the ⇋c for an acid; we are told that at ⇋ [HA] = 0.94 M, whereas [A-] = 0.060 M. The dissociation can be rewritten as: HA → H+ + A-. The ratio of [A-] to [H+] is 1:1, so [H+] must also be 0.060 M at ⇋. It follows, then, that: → Ka = [A-][H+] / [HA]. → (0.060)(0.060) / 0.94 = 3.8 x 10^-3. B. Ka is a measure of acid strength. An acid with a high Ka is a strong acid, because its ⇋ position is farther to the right, meaning dissociation is more complete. The Ka of sulfurous acid = 1.7 x 10^-2, & HA = 3.8 x 10^-3, which is smaller, meaning HA is a weaker acid. C. Kb is a measure of strength of a base. Because [H+] & [OH-] are related by Kw = 1.0 x 10^-14, Kb can be easily calculated. → Kb = (1.0 x 10^-14) / (3.8 x 10^-3) = 2.6 x 10^-12. D. pH = -log[H+] → pH = - log(0.060) = 1.22. E. pKa = -log[Ka] → -log(3.8 x 10^-3) = 2.42.

Write equations expressing what happens to each of the following bases in aqueous solutions: A. LiOH. B. Ba(OH)2. C. NH3. D. NO2-.

A. LiOH → Li+ + OH-. B. Ba(OH)2 → Ba^2+ + 2 OH-. C. NH3 + H2O → NH4+ + OH-. D. NO2- + H2O → HNO2 + OH-.

Identify each of the following as an Arrhenius acid or base, Bronsted-Lowry acid or base, or Lewis acid or base. A. NaOH, in: NaOH → Na+ + OH-. B. HCl, in: HCl → H+ + Cl- C. NH3, in: NH3 + H+ → NH4+. D. NH4+, in: NH4+ → NH3- + H+. E. (CH3)3N, in: (CH3)3N: + BF3 → (CH3)3N:BF3. F. BF3 in the equation above.

A. NaOH is an Arrhenius base. B. HCl is an Arrhenius & Bronsted-Lowry acid. C. NH3 is a Bronsted-Lowry & Lewis base. D. NH4+ is an Arrhenius & Bronsted-Lowry acid. E. (CH3)3N: is a Lewis Base. F. BF3 is a Lewis Acid.

Write chemical equations describing the buffer activity that prevents drastic pH changes when A. a strong acid and B. a strong base is added.

A. Strong Acid: → Salt in buffer dissociates completely: → NaX ⇋ Na+ + X-. →Added strong acid dissociates completely: → HCl ⇋ H+ + Cl-. → Protons from the acid are absorbed by the strong conjugate base of the salt: → H+ + X- ⇋ HX. B. Strong Base. → Weak acid in buffer hardly dissociates: → HX ⇋ HX. → Added strong base dissociates completely: → NaOH ⇋ Na+ + OH-. → OH- is a strong base, which attracts protons from the weak acid: → OH- + HX ⇋ H2O + X-.

A certain aqueous solution has [OH-] = 6.2 x 10^-5 M. A. Calculate [H+]. B. Calculate the pH of solution. C. Is the solution acidic or basic?

A. Water is composed of hydronium & hydroxide ions, & the dissociation constant of water, Kw, is defined as: → [H+][OH-] = 1.10 x 10^-14 M. If [OH-] = 6.2 x 10^-5, then: → [H+] = Kw/[OH-] = [1.0 x 10^-14/6.2 x 10^-5] = 1.6 x 10^-10 M. B. The pH of a solution is a logarithmic measurement of [H+], which expresses the degree of acidity. pH is defined as -log[H+]. In this case: → pH = -log(1.6 x 10^-10) = 9.79. C. Since the pH is 9.79, the solution is basic.

What volume of a 3 M solution of NaOH is required to titrate 0.05 L of a 4 M solution of HCl to the equivalent point?

At the equivalence point: → (Normality)acid(Volume)acid=(Normality)base(Volume)base. → 4 M HCl = 4 N HCl. → 3 M NaOH = 3 N NaOH. Plugging into the equation: → (4)(0.05) = (3)(Vb). → Vb = 0.067 L.

Element [102/20]Ω is formed as a result of 3 α and 2 β- decays. Which of the following is the parent element? A. [90/16]Γ. B. [114/24]Φ. C. [114/28]Θ. D. [112/8]Σ + [90/12]Π.

B. Emission of 3 α particles by the (unknown) parent results in the following changes: 1. Mass # = decreases by 3 x 4= 12 units. 2. Atomic # = decreases by 3 x 2 = 6 units. Emission of 2 β- particles results in: 1. Mass # = no change. 2. Atomic # = increases by 2 x 1 = 2 units. So the net change: 1. Mass # = (-12) + 0 = -12. 2. Atomic # = (-6) + (+2) = -4. Therefore, the mass number is 102 + 12, or 114. The atomic number is 20 + 4, or 24. The only possible answer, then, would be B.

Suppose a cobalt-60 nucleus undergoes β- decay such that: [60/27]Co → [A/Z+1]Y + β- What are A' and Z' of the daughter isotope?

Balance mass numbers. → 60 = A' + 0. → A' = 60. Then balance atomic numbers, taking into account that Cobalt has 27 protons, and 1 more proton on the right hand side. → Z' = Z + 1. → Z' = 27 + 1. → Z' = 28. Although the question does not ask for it, the daughter's identity is: [60/28]Ni.

If at time (t = 0) there is a 2-mole sample of radioactive isotopes of decay constant 2 (hour)^-1, how many nuclei remain after 45 minutes?

Because 45 minutes = 3/4 of an hour, the exponent (-λt): → -λt = -2(3/4) = -(6/4) = -(3/2). The exponential factor will be a number smaller than 1: → e^-λt = e^-(3/2) - 0.22. So 0.22 (or 22%) of the original 2-mole sample will remain. To find N°, multiply the # of moles by the # of particles per mole (Avogadro's Number): → N° = 2(6.02 x 10^23) = 1.2 x 10^24. From the equation that describes exponential decay, you can calculate the # of nuclei that remain after 45 minutes. → N = N°e^-λt. → N = (1.2 x 10^24)(0.22) = 2.6 x 10^23 particles.

If the half-life of a certain isotope is 4 years, what fraction of a sample of that isotope will remain after 12 years?

If 4 years is 1 half-life, then 12 years is 3 half-lives, so n=3. During the 1st half-life, or the 1st 4 years, half of the sample will have decayed. During the 2nd half-life, (years 4-8), half of what remains will decay, leaving 1/4 of the original. During the third, & final period (8-12), half of the remaining 1/4 will remain, meaning there will be 1/8 of the original sample. Thus, the fraction remaining after 3 half-lives is (1/2)^3, or 1/8.

For a certain [HA], Kb[A-] = (2.22 x 10^-11). Calculate the pH of a 0.50 M solution of HA.

If Kb = (2.22 x 10^-11), then: → Ka = [Kw / Kb] → [1.0 x 10^-14] / [2.22 x 10^-11]. → Ka = 4.5 x 10^-4. If HA dissociates according to the following expression: → HA ⇋ H+ + A- , then the ⇋ expression would be: → [H+][A-] / [HA]. We can let [H+] = x at ⇋, & since [H+] : [A-] = x. If the original [HA] was 0.5 M, at ⇋ [HA] = 0.5 - x. This expression then becomes: → 4.5 x 10^-4 = [x][x] / (0.5 - x). 0.5 - x = 0.5; since HA has a small Ka, its a weak acid. → 4.5 x 10^-4 = x^2 / 0.5. → x^2 = 2.25 x 10^-4. → x = 0.015 = [H+]. pH = -log[0.015] = 1.82.

Identify the conjugate acids & bases in the following equation: NH3 + H2O ⇋ NH4+ + OH-.

NH4+ is the conjugate acid of the weak base, NH3: OH- is the conjugate base of the weak acid, H2O.

A certain buffer solution is 3 M in HF & 2 M in NaF. Calculate the pH of this buffer given that the Ka of HF = 7.0 x 10^-4.

See the attached image for the work.

Suppose the isotope [A/Z]X emits a β+ and turns into an excited state of isotope [A'/Z']Y*, which then Y-decays to [An/Zn]Y, which in turn turn α-decays to [Am/Zm]W. If W is Fe-60 (Z=26), what is [A/Z]X?

Since the final daughter in this chain of decay is given, we can work backwards. First, the α-decay: → [An/Zn]Y → [60/26]Fe + [4/2]He^2+ Balancing atomic numbers: → Z" = 26 + 2 = 28. Balancing mass numbers: → A" = 60 + 4 = 64. So the Y-decay was: → [64/28]Y* → [64/28]Y + Y. The first reaction was β+ decay, so: → [A/Z]X → [64/28]Y* + β-. Again, balance the numbers. Z = 28 + 1 = 29. A = A + 0 = 64. The question did not ask, but you can to look at the periodic table to find the total chain, as attached.

Suppose a parent X alpha decays into a daughter Y such that: [238/92]X → [A'/Z']Y + α. What are the mass number (A') and atomic number (Z') of the daughter isotope Y?

Since α= [4/2]He, balancing the mass numbers & atomic numbers is all that needs to be done. → 238 = A' + 4. → A' = 234. → 92= Z' + 2. → Z' = 90.

A fission reaction occurs when Uranium-235 absorbs a low-energy neutron, briefly forming an excited state of Uranium-236, which then splits into Xenon-140, Strontium-94, and x more neutrons. How many neutrons are produced in the reaction? Note: the reaction in isotopic notation is found attached.

This question is asking "what is x?" By treating each arrow as an = sign, the problem is asking to balance the last "equation." The mass #s (A) on either side of each arrow must be =. This is an application of nucleon or baryon number conservation, which says that the total # of neutrons + protons remains the same, even if neutrons are converted to protons & vice versa, as they are in some decays. Because 235 + 1 = 236, the 1st arrow is balanced. To find the # of neutrons, solve for x in the last equation. → 236 = 140 + 94 + x. → x = 236 - 140 - 94. → x = 2. Therefore, 2 neutrons are produced in the rxn. These neutrons are free to go on & be absorbed by more U-235 & cause more fission, & the process continues.

Calculate the concentration of [H+] in a 2.0 M aqueous solution of acetic acid, CH3COOH (Ka= 1.8 x 10^-5).

Write out the ⇋ rxn: → CH3COOH (aq) ⇋ H+ (aq) + CH3COO- (aq). Write out the expression for acid dissociation constant. → Ka = [H+][CH3COO-] / [CH3COOH] = 1.8 x 10^-5. Since its a weak acid, [CH3COOH] at ⇋ is equal to its initial concentration, 2.0 M, less amount dissociated, x. [H+] = [CH3COO-] = x, since each molecule of CH3COOH dissociates into 1 [H+] & 1 [CH3COO-]. Thus, the equation can be written: → Ka = [x][x] / [2.0 - x] = 1.8 x 10^-5. We can approximate that [2.0 - x] ≈ 2.0 since acetic acid is a weak acid & x will be very small. Therefore: → Ka = [x][x] / [2.0] = 1.8 x 10^-5. → [x^2 / 2] = 1.8 x 10^-5. → 3.6 x 10^-5 = x^2. → x = 6 x 10^-3 M.

Element X is radioactive & decays via α-decay with a half-life of 4 days. If 12.5% of an original sample of Element X remains after t days, then determine t.

t = 12 days. Because the half-life element of X = 4 days, 50% of the original sample remains after 4 days, 25% remains after 8 days, and 12.5% remains after 12 days. Thus, t = 12 days. A different approach to this is to set (1/2)^n = 0/125, where n is the number of half-lives that have elapsed. Solving for n gives n=3. Thus, 3 half-lives have elapsed. Given the half-life is 4 days, t = 12 days.

If 10 mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added?

→ Add enough HCl to neutralize the solution. → Both acid & base = 1 M. → 10 mL of HCl will neutralize 10 mL of NaOH from Na*Va = Nb*Vb. → This produces 20 ml of 0.5 M NaCl solution. → Now calculate how much more HCl must be added to produce [H+] = 1 x 10^-2. → Let x be amount of HCl added. The total V of solution will be (20 + x) mL. Since this is a dilution problem, the amount of HCl added can be found by using the formula: → M1V1 = M2V2. → (1 M)(x mL) = (0.01 M)(20 + x) mL. → x = 0.2 mL, so a total of 10.2 mL of HCl is added to the original NaOH solution.

What is the mass defect & binding energy of this nucleus? Measurements of the atomic mass of a neutron & a proton yield: → Proton = 1.00728 amu. → Neutron = 1.00867 amu. A measurement of the atomic mass of 4He nucleus yields: → 4He = 4.00260 amu. 4He consists of 2 protons & 2 neutrons, which should theoretically give 4He a mass of: → 2(1.00728) + 2(1.00867) = 4.03190 amu.

→ The difference 4.03190 - 4.00260 = 0.02930 amu as a mass defect for 4He, & is interpreted as the conversion of mass into the binding energy of the nucleus. The rest energy (energy equivalent of a given mass x the speed of light squared) of 1 amu is 932 MeV, so using E=mc^2, we find that c^2= 932 MeV/amu. Therefore, the binding energy (BE) of 4He is: → BE = Δmc^2. → BE = (0.02930 amu)(932 MeV/amu). → BE = 27.3 MeV.