A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH2COONa) into 100.00 mL...

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Question

Answer 1

Answer #1

No.of milli moles of salt (sodium acetate) = (1 g) / (molar mass) x 1000

= (1 g) / (82.0343 g/mol) x 1000

= 12.19 milli moles

No.of milli moles of acid (acetic acid) = 100 mL x 0.100 M = 10 milli moles

No.of milli moles of hcl added = 1.80 mL x 10 M = 18 milli moles

Hence, Excess moles of acid = 18 - 12.19 = 5.81 milli moles

Total no.of moles of H⁺ in the solution = 5.81 + 10 = 15.81 milli moles

Concentration of H⁺ in the solution = (15.81) / (100 + 1.8 mL) = 0.1553 M

Therefore, pH of the solution will be :

pH = -log [H⁺]

pH = -log (0.1553)

pH = -(-0.809)

pH = 0.809 ------------ (**Answer**)

(in case of anything wrong/have any doubts, please reach out to me via comments. I will help you)

Answer 2

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