ACIDS AND BASES The equilibrium. You can leave out water itself of two aqueous s

Question

ACIDS AND BASES The equilibrium. You can leave out water itself of two aqueous solutions are descnibed in the table below. For each solution, write the chemical formulas of the major species present at Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases row, and the formulas of the species that will act as neither acids nor bases in the 'other row You will find it useful to keep in mind that HCH CO, is a weak acid. acids: 0.56 mol of KOH is added to 10 L of a 1.3 MHCH,CO,bases: D solution other: 55 mol of HCI is added to acids: 1.0 L of a solution that is 2 and KCH,Co a other: 0

Explanation / Answer

CH3COOH is weak acid. Moles of acetic acid = molarity* volume inL=1.3*1=1.3 moles

Moles of KOH= 0.56 moles, the reaction between acetic acid and KOH is

CH3COOH+ KOH-------àCH3COOK+ H2O

Theoretical molar ratio of CH3COOH: KOH= 1:1

Actual molar ratio of CH3COOH: KOH= 1.3:0.56 or 1.3/0.56:0.56/0.56= 2.3:1

So excess is CH3COOH and it remains after the reaction. Since CH3COOH remains, the solution will be acidic.

When 0.55 moles of HCl is added to mixture of CH3COOH and KOH.

Moles of acetic acid = moles of CH3COOK= 1.5*1= 1.5 moles

CH3COOK+HCl ------>CH3COOH+HCl,

The potassium acetate reacts with HCl and forms more acetic acid. Since this increases the concentration of CH3COOH and decreases the concentration of CH3COOK. Hence the solution will be acidic. Since pH= pKa+ log [A-]/[HA], [A-] is conjugate base at lower concentration and HA is acid. Hence pH will be less than previous case and the solution will be acidic.

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