Question

In: Chemistry

What is the pH of a buffer solution that is composed of a weak acid, HA...

What is the pH of a buffer solution that is composed of a weak acid, HA (Ka=8.02×10–9), and the conjugate base, A, after 3.24 mL of 0.089 M HCl solution is added. The initial concentrations of the 142 mL buffer solution are [HA]=0.58 M and [A]=0.61 M. Enter your value to two (2) decimal places.

Solutions

Expert Solution

pH of buffer solutions can be calculated by using Handerson's equation.

pH=pKa+log[salt]/[acid]

HCl is a strong acid.When HCl is added, it reacts with A- and forms:

                                               HCl    +    A-                   HA + Cl-

number moles of each substance can be calculated by using the formula: Molarity x volume in L

                                  HCl   +    A-                  HA                         +         Cl-

at t=0,           0.0324L x 0.089 M                 0.142L x 0.61M                        0.58M x 0.142L                    0

                    = 0.000288mol                              =0.0866mol                              =0.0824mol                       0

after reaction         0                                      0.0863mol    0.0827mol 0.000288mol

hence pH =      -log(8.02×10–9) + log( 0.0863mol)/(0.0827mol)

                = 8.1143

                =8.11


Related Solutions

consider a 1.00 M solution of a weak acid, HA. The pH of the solution is...
consider a 1.00 M solution of a weak acid, HA. The pH of the solution is found to be 3.85. A) calculate the [H3O+] in the solution. This would be the equilibrium concentration of H3O+ in the solution. B) write out an ICE table as before. Here, we don’t know the numerical value of Ka but we know the [H3O+] at equilibrium which you should see from your ICE table easily relates to the value of “x” in your table...
A 0.30 solution of a weak acid(HA) has a pH of 3.77. What is the Ka?
A 0.30 solution of a weak acid(HA) has a pH of 3.77. What is the Ka?
A buffer solution contains both a weak acid (HA) and its conjugate base (A-) so buffers...
A buffer solution contains both a weak acid (HA) and its conjugate base (A-) so buffers do not have large changes in pH after addition of a small amount of strong acid or strong base. What is the pH of 500.0 mL of a buffer composed of acetic acid (0.300 moles) and sodium acetate (0.200 moles) before and after addition of solid sodium hydroxide (5.84 g)? Assume no change in volume with addition of NaOH. pKa = 4.756 for acetic...
A student must make a buffer solution with a pH of 1.00. Determine which weak acid...
A student must make a buffer solution with a pH of 1.00. Determine which weak acid is the best option to make a buffer at the specified pH. formic acid,Ka = 1.77 x 10−4, 2.00 M sodium bisulfate monohydrate, Ka = 1.20 x 10−2, 3.00 M acetic acid, Ka = 1.75 x 10−5, 5.00 M propionic acid, Ka =1.34 x 10−5, 3.00 M Determine which conjugate base is the best option to make a buffer at the specified pH. sodium...
Buffer 1 is created such that the weak acid, HA, is 0.80 M and base, A–...
Buffer 1 is created such that the weak acid, HA, is 0.80 M and base, A– , is 0.80 M. Buffer 2 is created such that the weak acid, HA, is 0.080 M, and base, A– , is 0.080M. Which buffer has the greater buffer capacity? Briefly explain
A weak acid (HA) has a pKa of 4.009. If a solution of this acid has...
A weak acid (HA) has a pKa of 4.009. If a solution of this acid has a pH of 4.024, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)
A 0.20 M solution of the weak acid HA is 5.5 % ionized HA + H2O...
A 0.20 M solution of the weak acid HA is 5.5 % ionized HA + H2O --> H3O+ + A- a.Calculate the [H3O+] b.Calculate the [A- ] c.Calculate the [HA] at equilibrium d.Calculate the Ka for the acid e.Calculate the pH of the solution
4) In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what...
4) In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 49.5 mL of NaOH? Ka = 1.80 x 10-5 for HA. (3 significant figures) **Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4** 5) In a titration of a 100.0mL 1.00M HCl strong acid solution with...
1.What is the pH at the equivalence point when 0.100 M weak acid HA (Ka =...
1.What is the pH at the equivalence point when 0.100 M weak acid HA (Ka = 1.48 x 10 - 4) is titrated with 0.050 M KOH? 2.Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M KCN with 4.33 mL of 0.425 M strong acid HCl. HCN pKa = 9.21 3.Strong bases, NaOH and KOH can not be used as primary standards because of the presence of ______________ A.carbonate B.water C.adsorbed water please explain each...
A weak acid was HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a...
A weak acid was HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a volume of 100ml and a molarity of 0.100M. Find the pH at the following volumes of base added Vb= 0, 1, 5, 9, 9.9, 10, 10.1 and 12.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT