Question

1) A 42.0 mL sample of 0.108 M HNO2 is titrated with 0.258 M KOH. (Ka for HNO2 is 4.57×10−4.) Determine the pH at the equivalence point for the titration of HNO2 and KOH.

2) Two 27.0 mL samples, one 0.100 M HCl and the other 0.100 M HF, were titrated with 0.200 M KOH. Answer each of the following questions regarding these two titrations.

Part A

What is the volume of added base at the equivalence point for HCl?

Part B

What is the volume of added base at the equivalence point for HF?

Part C

Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral:

neutral for HF, and basic for HCl

neutral for HCl, and basic for HF

neutral for HCl, and acidic for HF

neutral for HF, and acidic for HCl

neutral for both

Part D

Predict which titration curve will have the lower initial pH.

HCl curve

HF curve

Answer 1

1) At equivalence point there is just a solution of potassium nitrite. The volume of base required for neutralization is found by first calculating the no.of moles of 0.108M nitrous acid in 42mL which is (42x0.108mM)/1mL = 4.536mmoles. Now, the volume of 0.258M KOH is found by equating the concentration and volume of KOH with the no.of moles of nitrous acid as 4.536mmoles = 0.258M x VmL of KOH = 17.5814mL. This gives the total volume of solution to now be 42 + 17.5814mL = 59.5814mL. So now, the concentration of the salt is not the same 0.108M as the volume has increased. We already found that there is 4.536mmoles of the salt. Thus, the cocentration in 59.5814mL is given by No.of moles of salt x (1000/volume) = 4.536mmoles x (1000/59.5814) = 0.07613M of salt.

Now, we calculate the pKb of nitrite ion from the pKa of nitrous acid as pKb = 14 - pKa. pKa = -logKa = 3.3401 which gives pKb = 10.6599.

As will be elucidated in the next question, the salt is a product of strong base and weak acid, making it basic in solution. So, to find out the concentration of hydroxide ions, Kb was found. Now, to find concentration of hydroxide ions, we take the square root of the product of pKb and concentration of the salt which gives sq.rt of 10^-10.6599x0.07613 = 1.2907x10^-6.

From this, pOH is found as pOH = -log[OH^-] = -log(1.2907x10^-6) = 5.88917 and from that, pH is found as pH = 14 - pOH = 8.1108. So at the equivalence point, the pH will be 8.1108.

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2) PART A. If 0.2M KOH is used to titrate samples of 0.1M HCl, both being strong acid and base, will conform with the law of equivalence as V_aM_a = V_bM_b from which we get V_b = (V_aM_a)/M_b = (27x0.1)/0.2 = 13.5mL of KOH.

PART B. HF is a weak acid with pKa = 3.17. At equivalence point, all of the acid should be neutralized. So, to find the quantity of HF in 27mL we get for 27mL of 0.1M HF (27 x 0.1mM)/1mL = 2.7mmoles of HF. To obtain the volume of 0.2M KOH required, we use 2.7mmoles of HF = 0.2M KOH x VmL of KOH = 13.5mL of KOH.

PART C. OPTION B. At the equivalence point in the titration with HCl, KCl and water will be formed. KCl is the salt of highly stable conjugate acid and base and so it will not hydrolyse water to form a more stable conjugate thus making the titration's equivalence point neutral. In the titration with HF, the salt formed is KF which is the salt of a strong base and weak acid. This shows that the conjugate acid of the base is quite stable while the conjugate base part is not. Therefore, the fluoride anion formed by dissolution of NaF in water stabilizes itself by forming HF in the medium. This HF is formed by abstraction of protons from water, releasing hydroxide ions,, making the medium basic. So in the titration with HCl, the pH will be neutral at equivalence and for the titration with HF, it will be slightly basic.

PART D. HCl is a strong acid and so it will ionize completely in water while HF will not ionize completely. Thus, the HCl titration curve will have a lower initial pH