Balanced chemical equations The value of Ka for benzoic acid , C6H5COOH , is 6.3

Question

Balanced chemical equations

The value of Ka for benzoic acid , C6H5COOH , is 6.30×10-5 . Write the equation for the reaction that goes with this equilibrium constant. (Use H3O+ instead of H+.) _____+_____=_______+_______

The value of Ka for acetic acid , CH3COOH , is 1.80×10-5 . Write the equation for the reaction that goes with this equilibrium constant. (Use H3O+ instead of H+.) _____+_____=_____+_______

The value of Ka for acetylsalicylic acid (aspirin) , HC9H7O4 , is 3.00×10-4 . Write the equation for the reaction that goes with this equilibrium constant. (Use H3O+ instead of H+.) _____+____=____+_____

Explanation / Answer

Answer – We are given the weak acids and need to write the balance the chemical equation at equilibrium. We know acid gives proton and from the conjugate base as follow -

For benzoic acid , C6H5COOH –

C6H5COOH + H2O = H3O+ + C6H5COO-

For acetic acid , CH3COOH-

CH3COOH + H2O = H3O+ + CH3COO-

for acetylsalicylic acid (aspirin) , HC9H7O4

HC9H7O4+ H2O = H3O+ + C9H7O4-

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