Answer to Question #131720 in General Chemistry for Anon

Question

2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of
water, which is carefully transferred to a conical flask. Then 100.00 mL of 0.120 M HCℓ is
added dropwise to this solution from a burette.

Answer 1

2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of

water .

Molar Mass of KCH3(CH2)2CO2  = 126 g / mol .

So , moles of KCH3(CH2)2CO2  = 2.643 / 126 = 0.02 mol ..

Given moles of HCl = 100 * 0.120 / 1000 = 0.0120 mol .

When HCl React with  potassium butanoate then   Butanoic Acid is formed .

So Hcl Is limiting Reagent( in lower amount ) So HCl Totally Consumed and  potassium butanoate and   Butanoic Acid Are Formed . this is buffer solution .

Moles of potassium butanoate =(0.02 - 0.0120) mol . = 0.008

Moles of butanoic acid = (0.0120) mol .

So , pH = pKa of butanoic acid + log [  potassium butanoate ] / [ Butanoic Acid ].

pH = 4.824 + log 0.008 / 0.0120 = 4.824 -0.174 = 4.65 Answer .