(3) 7. An acid HA is dissolved in water: HA H++ A. [HA] 0.250 Mi [H -4.00 x 10-M

ID: 1038979 • Letter: #

Question

(3) 7. An acid HA is dissolved in water: HA H++ A. [HA] 0.250 Mi [H -4.00 x 10-M; [A1-4.00 x 104 M Calculate the pka for the acid. Show your work below. (2) 8. write the balanced equation for the neutralization reaction between HI and Ba(OH)2 n aqueous solution Include phases (physical states) in your equation. (2) 9. Write the balanced equation for the reaction of an aqueous solution of HBr (hydrobromic acid) and aluminum metal. Include physical states in your equation. 2Al (s) + 6HBr (aq)?? ? 2A1Br3 (aq) + 3H2 (g) . (6) 10. Fill in the blanks in the table below for three aqueous solutions. Acid Soln. H IOH] 0.00535 M 2.10 x 10- M 9.25 (3) 11. In a titration it is found that 97.7 mL of 0.154 MNaOHag) is needed to neutralize 25.0 mL. of a solution of HCleg). Calculate the concentration of the HCl solution. Show your work. [Hint: Write a balanced equation for the reaction first.]

Explanation / Answer

7: pKa value is used to express the acidity of weak acids, It can be very large or very small. For example, the Ka value for acetic acid (CH3COOH) is 0.0000158, while the Ka value for ascorbic acid is 1.6 X 10-12. Working with such numbers is inconvenient, so to makes things easier, chemists have defined the pKa number as:

pKa = -log Ka

According to this definition, the pKa value for acetic acid is 4.8, while the pKa for ascorbic acid is 11.80. The smaller the pKa number, the stronger the acid.

pKa can be calculated using given formula

pKa= -log10 Ka

where Ka = [A-] [H+]/[AH]

[4.00x10-4] X [4.00x10-4]/0.250M=6.4x10-7

Ka=6.4x10-8

=-log10 (6.4 x10-8)

pKa=7.19

8: Neutralization reactions

2 HI(aq) + BA(OH)2(aq) = 2 H2O (l) + BAI2(s)

It’s a double replacement reaction

9: 2 Al(s) + 6 HBr(aq) = 2 AlBr3 (s) + 3 H2 (g)

It’s a single replacement reaction

10: H+, OH-and PH can be calculated using given formulas

Solution

H+

OH-

0.00535 M

2.275

1.88 x10-12 M

0.00066 M

7.322

2.10 x10-7 M

5.6 x10-10 M

9.25

1.77 x10-5 M

PH=-log [H+]

[H+]=10-PH

POH=-log [OH-]

[OH-]=10-POH

PH + POH =14

11: At the equivalence point in neutralization, the moles of acid are equal to the moles of base, assuming they react in a 1:1 ratio according to the balanced neutralization equation.

moles of acid = moles of base

molarity of the NaOH solution = 0.154 M= MB

volume of the NaOH solution = 97.7 mL=VB

volume of the HCl solution = 25 mL=VA

molarity of the HCl solution = ? MA

NaOH + HCl = NaCl + H2O.

The given equation can be used to determine the molarity of the acid.

MA × VA = MB × VB

MA is the molarity of the acid, while MB is the molarity of the base. VA and VB are the volumes of the acid and base, respectively.

MA= MB x VB/VA

=0.154x 97.7/25=0.601M