20 mol of M//10 CH(3)COOH solution is ti...

`20 mol` of `M//10 CH_(3)COOH` solution is titrated with `M//10 NaOH` solution. After addition of `16 mL` solution of `NaOH`. What is the `pH` of the solution `(pK_(a) = 4.74)`

4.858

8.588

3.858

2.558

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CH_(3)COOH is titrated with NaOH solution. Which is true statement?

Calculate the pH of the solutions when following conditions are provided: a. 20mL of M//10 CH_(3)COOH solution is titrated with M//10 solution of NaOH . i. No titration is carried out. ii. When 10mL of NaOH is added. iii. When 20mL of NaOH is added. iv. When 30mL of NaOH is added. (pK_(a) of CH_(3) COOH = 4.74) b. 20mL of M//10 NaOH solutions sio titrated with M//10 solution of CH_(3)COOH . i. No titration is carried out. ii. When 18mL of Ch_(3)COOH is added. iii. When 20mL of CH_(3)COOH is added. iv. When 40mL of CH_(3)COOH is added. c. 10mL of M//10 NH_(4)OH solution is titrated with M//10 solution of H_(2)SO_(4) . i. No titration is carried out. ii. When 4mL of H_(2)SO_(4) is added. iii. When 5mL of H_(2)SO_(4) is added. iv. When 10mL of H_(2)So_(4) is added. pK_(a) of NH_(4)OH = 4.76 d. 10mL of M//10H_(2)SO_(4) solution is titrated with M//10 solution of NH_(4)OH . i. No titration is carried out. ii. When 10mL of NH_(4)OH is added. iii. When 20mL of NH_(4)OH is added. When 40mL of NH_(4)OH is added.

10 mL of N//20 NaOH solution is mixed with 20 mL of N//20 HCI solution. The resulting solution will :

100 ml of 0.2 M CH_3COOH is titrated with 0.2 M NaOH solution. The pH of of the solution at equivalent point will be ( pKa of CH_3COOH=4.76 )

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MOTION-IONIC EQUILIBRUIM-Exercise -2 (Level -II)

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