Chapter 16 A

Calculate pH for 1.6×10−3 M Sr(OH)2.

pH = 14 - pOH pOH = -log(OH) = 2.49 pH = 14-2.49 = 11.51

The average pH of normal arterial blood is 7.40. At normal body temperature (37∘C), Kw=2.4×10−14. Calculate the pOH

pOH = -log (OH) = -log (6.0 x 10^-7) = 6.22

pOH Equation

pOH=-log[OH-]

pH + pOH

14.00

pH Equation

pH = -log[H+]

The average pH of normal arterial blood is 7.40. At normal body temperature (37∘C), Kw=2.4×10−14. What's the difference in pH values of the two solutions?

ΔpH = - log ([H+]B/ 250 [H+]B = - log (1 / 250) = 2.4

NH3(g) and HCl(g) react to form the ionic solid NH4Cl(s) Which substance is the Brønsted-Lowry acid in this reaction? Which is the Brønsted-Lowry base?

HCl is the Bronsted-Lowry Acid; it donates an H+ to NH3+ to form NH4; NH3 is the Bronsted-Lowry Base; it accepts an H+ from HCL

Give the conjugate base of the following Bronsted-Lowry acid: HIO3.

IO3^- A conjugate base has one less H+ than it's conjugate acid

If you mix equal concentrations of HX and HY, will the equilibrium HX(aq)+Y−(aq)⇌HY(aq)+X−(aq) lie mostly to the right (Kc> 1) or to the left (Kc< 1)?

It will lie mostly to the left (Kc<1) Equilibrium reactions will favor the side with the weaker acid-base pair In this case, the weaker pair is HX and Y−. These species are located on the reactants side of the equation, so the equilibrium lies to the left.

Which of these statements about how the percent ionization of a weak acid depends on acid concentration is true?

Line C is the most accurate, b/c as the acid concentration increases, the percent ionization decreases

Give the conjugate acid of the following Bronsted-Lowry base: O2−.

OH^- A conjugate acid has one more H+ than it's conjugate base

Calculate [OH−] for 1.6×10−3 M Sr(OH)2.

SrOH2 is a strong base and will completely dissociate in solution. Therefore, the [OH−] is based on the chemical formula of the compound; there are two OH− ions for every Sr(OH)2 molecule that dissociates. [OH−]=2[Sr(OH)2] = 2(1.6×10−3 M) = 3.2×10−3 M OH−

The following diagrams(Figure 1) represent aqueous solutions of two monoprotic acids, HA (A=X or Y). The water molecules have been omitted for clarity. With HX, there are four HA(acid) molecules, two H3O+ molecules, and two A- ions present. With HY, there are two HA(acid) molecules, four H3O+ molecules, and four A- ions present. Which is the stronger acid, HX or HY?

The Stronger Acid is HY The stronger acid will dissociate to a greater extent Each start with acid molecules apiece; HY has dissociated while HX only has two

Define Kw

The ionic product of water Kw = [H+][OH-] = 1.0 x 10^-14 For pure water, Kw = [H+]^2

At 50 ∘C, the ion-product constant for H2O has the value Kw=5.48×10−14. Based on the change in Kw with temperature, predict whether ΔH is positive, negative, or zero for the autoionization reaction of water: 2H2O(l)⇌H3O+(aq)+OH−(aq)

The value of Kw increases with increasing temperature, so the ΔH is positive The autoionization of water is endothermic

With HX, there are four HA(acid) molecules, two H3O+ molecules, and two A- ions present. With HY, there are two HA(acid) molecules, four H3O+ molecules, and four A- ions present. Which is the stronger base, X− or Y−?

X- is a stronger base Since HY is the stronger acid it forms the weaker conjugate base

At 50 ∘C, the ion-product constant for H2O has the value Kw=5.48×10−14. What is the pH of pure water at 50 ∘C?

[H+[^2 = 5.8x10^-14 [H+] = 2.41 x 10^-7 pH = -log (2.41 x 10^-7) = 6.62

The average pH of normal arterial blood is 7.40. At normal body temperature (37∘C), Kw=2.4×10−14. Calculate the concentrations of [H+] and [OH-]

[H+] = 10^-pH = 10^-7.40 = 4.0 x 10^ -8 [OH-] = kw/[H+] = (2.4 x 10^-14) / (4.0 x 10^8) = 6.0 x 10^-7

If a neutral solution of water, with pH = 7.00, is cooled to 10 ∘C, the pH rises to 7.27.

[H+] = [OH-] In a neutral solution, the concentrations will decrease but are still equal at 10^-7.27

Percent Ionization Equation

[H+] equilibrium /[HA] initial x 100%