Part A: what is the pKa and Ka of this acid? (Y=7.5 x 10^-5X + 1.5 x10^-4 Part B

Question

Part A: what is the pKa and Ka of this acid? (Y=7.5 x 10^-5X + 1.5 x10^-4
Part B: a monoprotic acid, 0.215 g, was tirated with 32.3 mL of 0.156 M NaOH to its endpoint. What is this acids molar mass? Part A: what is the pKa and Ka of this acid? (Y=7.5 x 10^-5X + 1.5 x10^-4
Part B: a monoprotic acid, 0.215 g, was tirated with 32.3 mL of 0.156 M NaOH to its endpoint. What is this acids molar mass?
Part B: a monoprotic acid, 0.215 g, was tirated with 32.3 mL of 0.156 M NaOH to its endpoint. What is this acids molar mass?

Explanation / Answer

part A) for any diprotic acid H2A with ka1,ka2 as the acid dissociation constants for H2A and HA- forms,

ka=ka2/ka1=7.5 x 10^-5/1.5 x10^-4=5*0.1=0.5

pka=-log ka=-log 0.5=0.301

partB)moles of NaOH used =volume*molarity=0.0323L*0.156mol/L=0.00504 moles

A monoprotic acid reacts with NaOH in the molar ratio of 1:1

moles of NaOH used=moles of acid=0.00504mol

moles of acid=mass of acid/molar mass=0.00504mol

or,molar mass of acid=mass of acid/0.00504mol=0.215g/0.00504mol=42.659g/mol

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