Concept explainers
Sulfanilic acid
A buffer is prepared by dissolving 0.20 mol of sulfanilicacid and 0.13 mol of sodium sulfanilate
- Compute the pH of the solution.
- Suppose 0.040 mol of HCl is added to the buffer.Calculate the pH of the solution that results.
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Chapter 15 Solutions
Principles of Modern Chemistry
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- What is the equilibrium expression for: 2NH3 ⇌ N2 + 3H2arrow_forwardThe equilibrium below is a key step in the ocean acidification process. CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3− (aq) ΔH < 0 What is the effect of increasing the temperature of this equilibrium? Group of answer choices The pH will increase The [H+] will increase CO2 pressure will decrease No change will be observedarrow_forwardWater ionizes by the equation H₂O(1) H+ (aq) + OH- (aq) The extent of the reaction is small in pure water and dilute aqueous solutions. This reaction creates the following relationship between [H+] and [OH-]: Kw = [H+][OH-] Keep in mind that, like all equilibrium constants, the value of Kw changes with temperature.arrow_forward
- Write the equilibrium constant expression for this reaction: NH(aq) → NH3(aq) +H* (aq) X 믐 5arrow_forwardWrite the formula for the K eq for the following reaction: N 2(g) +3H 2(g) ->2NH3(g)arrow_forwardConsider the following acidic equilibrium: H₂CO₃(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + H₃O⁺(aq). If you add NaHCO₃ to this solution, which of the following will occur? A) The reaction quotient will decrease. B) The reaction will shift in the reverse direction. C) The equilibrium constant will increase. D) No changes to the equilibrium positions will take place.arrow_forward
- consider the following equilibrium. CrO4(aq) + 2H(aq) <---> Cr2O7(aq) + H2O(l) (yellow) (Orange) If a strong acid, such as HCl, is added to the solution, in which direction will the equilibrium shift and what color will the resulting solution be.arrow_forwardAcetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (d) When starting with completely un-dissociated acetic acid, is it accurate to assume that [HOAc]0 = [HOAc]eq? Why or why not? (e) A highly concentrated acetic acid solution contains 15.0M acetic acid at equilibrium. What are the equilibrium concentrations of the hydronium and acetate ions in this solution? (f) Creating the concentrated acetic acid solution by dissolving liquid HOAc in water raises the temperature of the water by about 5°C from room temperature. At 50°C, do you expect the solution to contain more or less acetate ion OAc– than what you calculated in (c)? Why?arrow_forwardAcetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (a) Write the equilibrium constant expression for the dissociation of acetic acid. (b) Vinegar sold commercially is typically 0.8 − 1.0 M acetic acid. A 1.00 M solution of acetic acid is measured by its pH to have an equilibrium concentration of 4.19×10−3 M for both acetate ions and hydronium ions at room temperature. Assuming [HOAc]0 = 1.00M, what is the equilibrium concentration of undissociated acetic acid [HOAc]eq to the correct number of significant figures? (c) What is the value of the equilibrium constant Keq for the dissociation according to the concentrations from part (b)? (d) When starting with completely un-dissociated…arrow_forward
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