In this experiment, the weak acid (HA), is reacted with 0.100 M NaOH using a tit

Question

In this experiment, the weak acid (HA), is reacted with 0.100 M NaOH using a titration set-up. Write the chemical equation for the neutralization reaction. Calculate the moles of acid that would react with 30 mL of 0.100 M is referred to as the equivalence point of the titration. If the molar mass of the acid were 100 g/mol. how many grams o would react? How many moles of HA are present at point B (the half-equivalence point)? How many moles of HA are present at point C (the equivalence point)? The titration technique is used to do the added sequentially and the pH is monitored as the moles of HA react to form NaA. The Ka is calculated at the three lettered points. Use the information is question 1 to find: How many moles of HA are present at point A (initial)? How many moles of HA are present at point B (the half-equivalence point)? How many moles of HA are present at point C (the equivalence point)? In the equilibrium equation for the ionization of HF dissolved in water. HF + H_2O H_3O* + F^-. (point A) how are the concentrations of H_3O^+ and F^- obtained? At point B. the concentration of F^- would be equal to Write the equilibrium reaction for the hydrolysis of the F^- ion in NaF (point C). How is this equilibrium related to the Ka of the weak acid?

Explanation / Answer

Q1.

a)

For neutralization

NaOH(aq) + HA(aq) <--> H2O(l) + NaA(aq)

b)

moles of acid if

mol of base = MV = (30*10^-3)(0.1) = 0.003 mmol of NaOH used

so... assume 1:1 ratio

so

0.003 mol of acid HA were present

c)

molar mass of acid = 100 g/mol

so:

total mass of acid = mol*MW = (0.003)(100) = 0.3 grams of acid

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