Rainwater is acidic because CO21g2 dissolves in the water, creating carbonic acid, H2CO3. If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO3). Calculate the concentrations of carbonic acid, bicarbonate ion 1HCO3-2 and carbonate ion 1CO32 - 2 that are in a raindrop that has a pH of 5.60, assuming that the sum of all three species in the raindrop is 1.0 * 10-5 M.
Ch.17 - Additional Aspects of Aqueous Equilibria
17장, 문제 85
Furoic acid (HC5H3O3) has a Ka value of 6.76 × 10^-4 at 25 _x001F_C. Calculate the pH at 25 _x001F_C of (a) a solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate (NaC5H3O3) to enough water to form 0.250 L of solution; (b) a solution formed by mixing 30.0 mL of 0.250 M HC5H3O3 and 20.0 mL of 0.22 M NaC5H3O3 and diluting the total volume to 125 mL; (c) a solution prepared by adding 50.0 mL of 1.65 M NaOH solution to 0.500 L of 0.0850 M HC5H3O3.
검증된 단계별 안내1
Step 1: For part (a), calculate the molarity of furoic acid (HC5H3O3) and sodium furoate (NaC5H3O3) in the solution. Use the formula: Molarity = (mass of solute / molar mass of solute) / volume of solution in liters.
Step 2: Recognize that the solution in part (a) is a buffer solution. Use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (sodium furoate) and [HA] is the concentration of the acid (furoic acid).
Step 3: For part (b), calculate the initial moles of HC5H3O3 and NaC5H3O3 using their given concentrations and volumes. Then, find their concentrations in the final 125 mL solution.
Step 4: Again, use the Henderson-Hasselbalch equation for part (b) to find the pH, using the concentrations calculated in Step 3.
Step 5: For part (c), calculate the moles of NaOH added and determine the moles of HC5H3O3 remaining after the reaction with NaOH. Use the stoichiometry of the reaction to find the concentrations of the remaining acid and its conjugate base, then apply the Henderson-Hasselbalch equation to find the pH.
주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Acid-Base Equilibrium
Acid-base equilibrium refers to the balance between the concentrations of acids and their conjugate bases in a solution. The dissociation constant (Ka) quantifies the strength of an acid, indicating how well it donates protons (H+) in solution. Understanding this concept is crucial for calculating pH, as it helps determine the extent of dissociation of the acid and the resulting concentrations of H+ ions.
추천 영상:
가이드 코스
Triprotic Acid Equilibrium
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a mathematical formula used to calculate the pH of a buffer solution. It relates the pH to the pKa of the acid and the ratio of the concentrations of the conjugate base to the acid. This equation is particularly useful in scenarios involving weak acids and their salts, such as furoic acid and sodium furoate, allowing for straightforward pH calculations in buffered solutions.
추천 영상:
가이드 코스
Henderson-Hasselbalch Equation
Stoichiometry in Acid-Base Reactions
Stoichiometry involves the calculation of reactants and products in chemical reactions. In acid-base reactions, it is essential to understand the molar relationships between acids, bases, and their conjugate pairs. This concept is vital for determining the concentrations of species in solution after mixing, which directly impacts the pH calculations in the scenarios presented in the question.
추천 영상:
가이드 코스
Acid-Base Reaction
관련 실천
교과서 질문
789
views
교과서 질문
(b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group 3?
465
views
교과서 질문
Derive an equation similar to the Henderson–Hasselbalch equation relating the pOH of a buffer to the pKb of its base component.
910
views
교과서 질문
The acid–base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the pH is 4.68. What is the pKa for bromcresol green?
1392
views
