The following questions a24 - a26 relate to 100 ml of 0.0150 m solution of benzoic acid (c6h3cooh). ka(c6h3cooh) = 6.4 x 10^-5. what is the ph of the solution after the addition of 1 x 10^-3 moles of naoh? you may assume no volume change to the solution upon addition of the naoh.

PH of the weak acid: approximately 2.513.

PH of the buffer solution: approximately 4.495.

Explanation:

The Ka value of benzoic acid is much smaller than 1. Benzoic acid will dissociate but only partially when dissolved in water. Construct a RICE table for this process. Let the equilibrium of be . Note that .

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Solve this quadratic equation for :

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Take only the non-negative root. .

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Each benzoic acid contains only one carboxyl group . Benzoic acid is thus a monoprotic acid. Each mole of the acid will react with only one mole of . The 100 mL solution initially contains moles of benzoic acid. The moles of will neutralize only part of the acid. The solution will eventually contain moles of (from the salt ) and moles of .

Both the acid and the conjugate base of the acid exist in large amounts in the solution. Apply the Henderson-Hasselbalch equation for weak acid buffers to find the pH of this buffer solution.

for benzoic acid.

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answer: h+ its less than oh

explanation:

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.