In the neutralization reaction between the sodium hydroxide (NaOH) and acetic ac

Question

In the neutralization reaction between the sodium hydroxide (NaOH) and acetic acid (HC_2 H_3 O_2), it will take 1 mole of NaOH to neutralize every 1 mole of HC_2 H_3 O_2. Based on this experiment (32-33): Calculate the number of moles of HC_2 H_3 O_2 if 0.001 moles of NaOH is consumed. 0.001 moles 1.00 moles 0.10 moles 0.010 moles 0.20 moles Calculate the concentration (M) of the acetic acid (HC_2 H_3 O_2) solution if 0.010 L was used in the titration. Molarity = moles/Liters 0.001 M 1.00 M 0.125 M 0.091 M 0.100 M What is the pH of a solution with [H_3 O^+] = 1 times 10^-5 M? 1.0 times 10^-5 9.0 5.0 -5.0 -9.0

Explanation / Answer

The neutralization reaction between acetic acid and sodium hydroxide is

CH3COOH + NaOH -----> CH3COONa + H2O

here, 1 mol of acetic acid is neutralized by 1 mol of sodium hydroxide.

32)

Option [a]

If 0.001 mols of NaOH were reacted in neutralization reaction , then according to stoichiometry, to neutralize, the acetic acid consumed would be 0.001 mols

33)

Option [E]

The moles of acetic acid used in neutralization are 0.001 moles. and the volume of the acetic acid solution used is 0.01L.

Hence by definition of molarity (M) = no. of moles / litre of solution

molarity (M) = 0.001 mols / 0.01L = 0.1 M

Molarity (M) = 0.1M

33)

Option [C]

[H3O+] = 1*10-5 M

[H3O+] can also be stated as [H+] ions

[H+] = 1*10-5 M

we know,

pH = - log [H+]

pH = - log [1*10-5 ]

pH = 5

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