A solution was prepared by mixing 10.0 mL of 0.50 M NaOH with 10.0 mL of 1.00 M acetic…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
A solution was prepared by mixing 10.0 mL of 0.50 M NaOH with 10.0 mL of 1.00 M acetic acid, Ka=1.8 ×10–5. Find the pH of solution.
- 2.45
- 1.67
- 2.37
- 4.74
Answer
(D) 4.74
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A solution which is 10 –3 M each in Mn 2+ , Fe 2+ , Zn 2+ and Hg 2+ is treated with 10 –16 M sulphide ion. If…
- Which among the following represent the conjugate acid/base pairs ?
- At 90°C, pure water has [H 3 O + ] as 10 –6 mole litre –1 . What is the value of K w at 90°C?
- The species present in solution when CO 2 is dissolved in water are (2006)
- Which of the following solutions will have pH close to 1.0 ? (1992)
- The product of the concentrations of the ions of an electrolyte raised to power of their coefficients in the…
- Volume of the flask in which the following equilibria are separately established are transferred to a flask…
- When equal volumes of the following solutions are mixed, precipitation of AgCl ( K sp = 1.8 × 10 –10 ) will…