Interpretation:
To calculate the
Concept introduction:
Answer to Problem 43QAP
Explanation of Solution
At
For above reaction equilibrium constant K is:
Given,
The value of equilibrium constant
Hence,
Thus,
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Chapter 21 Solutions
Chemistry: Principles and Reactions
- The hydrogen phthalate ion, C8HsO4, is a weak acid with Ka = 3.91 106. C8H5O4(aq)+H2O(l)C8H4O42(aq)+H3O+(aq) What is the pH of a 0.050 M solution of potassium hydrogen phthalate. KC8H5O4? Note: To find the pH for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the H3O+ concentration is [H3O+]=Ka1Ka2 For phthalic acid, C8H6O4 is Ka1 is 1.12 103, and Ka2 is 3.91 106.arrow_forwardWhen carbon dioxide dissolves in water it reacts to produce carbonic acid, H2CO3(aq), which can ionize in two steps. H2CO3(aq)HCO3(aq)+H+(aq)Kc1=4.2107HCO3(aq)CO32(aq)+H+(aq)Kc2=4.81011 Calculate the equilibrium constant for the reaction H2CO3(aq)CO32(aq)+2H+(aq)arrow_forwardWhat is the pH of a 1.00 molar solution of NH4Cl(aq)? The Kb for NH3 = 1.8 × 10–5 .arrow_forward
- The pH of an aqueous solution of 8.39×10-2 M ammonium nitrate, NH4NO3 (aq), isarrow_forward8. (a) HA(aq) is a weak acid with a dissociation constant, Ka, of 8.8 x 10−12. What is the pH of a 0.022 M solution of A−(aq)? The temperature is 25 ◦C. (b) For the reaction A(g) =A(l), the equilibrium constant is 0.666 at 25.0 ◦C and 0.222 at 75.0 ◦C. Making the approximation that the entropy and enthalpy changes of this reaction do not change with temperature, at what temperature will the equilibrium constant be equal to 0.777?arrow_forwardWhat is the pH of a 1.00 molar solution of NaCN(aq)? The Ka for HCN = 6.2 × 10–10 .arrow_forward
- Calculate the equilibrium concentrations of all chemical species present in a 0.10 M H2CO3(aq) solution. What is the percent ionization of the acid and resulting pH?arrow_forwardHA(aq) is a weak acid with a dissociation constant, Ka, of 8.8 x 10−12. What is the pH of a 0.022 M solution of A−(aq)? The temperature is 25 ◦C.arrow_forwardCalculate the pH at 25 °C of a solution that is 1.00 M in NaClO(aq) and 2.00 M in HClO(aq). For HClO, Ka = 6.8×10−4 at 25 °C.arrow_forward
- Write the equilibrium constant expression, K, for the following reaction taking place in dilute aqueous solution.ClO- (aq) + H2O (l)HClO (aq) + OH- (aq)arrow_forwardThe pH of an aqueous solution of 0.400 M hydrocyanic acid, (Ka (HCN) = 4.00 × 10-10) isarrow_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_forward
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