What important information does the K, give? In this experiment, the weak acid (

Question

What important information does the K, give? 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 reaction. Calculate the moles of acid that would react with 30 mL of 0.100 M NaOH. This is referred to as the equivalence point of the titration. If the molar mass of the acid were 100 g/mol. how many grams of the weak acid would react? The titration technique is used to do the reaction because the 0.1 M NaOH can be 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. 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. how are the concentrations of H_3O^+ and F^-obtained? At point B. the concentration of F would be equal to

Explanation / Answer

Solution :-

Q1) Ka gives the information about the acid strength that if the ka is small then the acid is weak and if the ka is large then acid is strong

Q2)

a) Balanced equation for the reaction of HA with NaOH is as follows

HA + NaOH -------- > NaA + H2O

b) moles of NaOH = molarity * volume in liter

                                 = 0.100 mol per L * 0.030 L

                                 = 0.003 mol NaOH

Mole ratio of the HA to NaOH is 1 : 1 so the moles of HA reacted = 0.003 mol HA

c) mass of acid = moles * molar mass

                           = 0.003 mol * 100 g per mol

                           = 0.300 g acid HA

Q3)

a) Moles of HA present initially = 0.003 mol

b) Point B is the half equivalence point so the moles of HA present at point B = 0.003 / 2 = 0.0015 mol HA

c) At point C all the HA is reacted with NaOH so the moles of HA at point C= 0

Q4) in the equilibrium equation of the

HF + H2O ---- > H3O+ + F^-

The concentration of the H3O+ and A- is obtained by using the acid dissociation constant Ka

Q5) At point B the concentration of the F- = (0.0015 mol / 0.045 L) = 0.0333 M

Q6) a) hydrolysis of the F- in NaF

Na^+ + F^- + H2O ----- > HF + Na^+ + OH^-

b) This equilibrium is weak (lower) than the ka of weak acid

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