The pKa of certain weak acid is 4.0. What should be the salt-to-acid ratio if we have to prepare a buffer with pH = 3 using the acid and of its salts?
Correct answer: D. 1:10
- A. 4:5
- B. 10:1
- C. 5:4
- D. 1:10
Explanation
To prepare a buffer with a pH of 3 using a weak acid with a pKa of 4.0 and its corresponding salt, we need to consider the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) In this equation, [A-] represents the concentration of the conjugate base (salt) and [HA] represents the concentration of the weak acid. Given that we want a pH of 3, we substitute the values into the equation: 3 = 4.0 + log([A-]/[HA]) Simplifying the equation: -1 = log([A-]/[HA]) To obtain the salt-to-acid ratio, we need to convert the logarithmic equation into an exponential form. In this case, we can rewrite it as: [A-]/[HA] = 10^(-1) Taking the logarithm base 10 of both sides: log([A-]/[HA]) = -1 From this equation, we can conclude that the salt-to-acid ratio should be 1:10 (or [A-]/[HA] = 1/10) in order to achieve a buffer with a pH of 3 using the weak acid and its salt.
Last updated
About Chemical Equilibrium
Reversible reactions reach dynamic equilibrium when forward and reverse rates become equal, and Le Chatelier's principle predicts the effect of changing concentration, pressure or temperature. The chapter also covers solubility product, the common ion effect, buffer action and the conditions used in Haber's process, with equilibrium shifts distinguished from changes in the equilibrium constant.
Practise Chemical Equilibrium
625 free Chemical Equilibrium MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1.8 x 10 -5 is the dissociation constant of:
1 mole of CH3COOH and 1 mole of C2H5OH react to produce 2/3 mole of CH3COOC2H5. The equilibrium constant is
1 mole of ethyl alcohol was treated with one mole of acetic acid at 25°C. 2/3 of the acid changes into ester at equilibrium. The equilibrium constant of the reaction will be:
2HF ⇌ H2 + F2 PCI5 ⇌ PCI3 + Cl2The statement that is false about the Ke of both the reactions is;
2O3 ⇌ 3O2 Ke = 1055 at 25°C, it explains