pH with a pH meter and have your instructor record this value and sign your note

Question

pH with a pH meter and have your instructor record this value and sign your notebook. d) Combine all the solutions in one flask. How many moles of phosphoric acid are present in the flask? How many moles of NaoH are required to completely react with all the moles of phosphoric acid? What volume of 1M NaOH is required to completely neutralize (react with) all the moles of phosphoric acid? Measure out the required amount of 1 M NaOH and add it to the flask. Measure the pH and pour it down the drain if the ph is between 6-8.

Explanation / Answer

c) To prepare 100 ml of phosphoric acid solution with pH 5.5

pH = log[H+]

[H+] = 3.16 x 10^-6 M

H3PO4 <==> H+ + H2PO4-

[H+] = [H2PO4-]

Ka1 = [H+][H2PO4-]/[H3PO4]

feeding the values,

7.1 x 10^-3 = (3.16 x 10^-6)^2/[H3PO4 + 3.16 x 10^-6]

[H3PO4] = 1.41 x 10^-9 M + 3.16 x 10^-6

               = 3.16 x 10^-6 M

moles of H3PO4 to be taken = 3.16 x 10^-6 M x 0.1 L

                                              = 3.16 x 10^-7 mol

mass of H3PO4 to be taken = 3.16 x 10^-7 mol x 98 g/mol

                                             = 3.1 x 10^-5 g

                                             = 0.031 mg

So we dissolve 0.031 mg of H3PO4 in 100 ml water to get a solution of pH 5.5

d) moles of NaOH needed for neutralization = moles of H3PO4 present

                                                                      = 3.16 x 10^-7 mol

Volume of NaOH required = moles/molarity

                                           = 3.16 x 10^-7 mol x 1000/1 M

                                           = 3.16 x 10^-4 ml

3.

a) To prepare 100 ml of 1 M H3PO4 solution

molarity = moles/volume

moles = mass/molar mass

1 = mass of H3PO4/98 x 0.1

So,

mass of H3PO4 to be taken = 9.8 g

pH calculation,

H3PO4 + H2O <==> H3O+ + H2PO4-

let x amount has dissociated

7.1 x 10^-3 = x^2/1

x = [H3O+] = 0.0843 M

pH = -log[H3O+] = 1.07

b) To prepare a 100 ml of 30 mM (0.03M) solution,

Volume of 1 M solution to be taken = final molarity x final volume/initial molarity

                                                        = 0.03 M x 100 ml/1 M

                                                        = 3 ml

so 3 ml of 1 M H3PO4 solution diluted to 100 ml would give a 30 mM solution

pH of solution,

with x amount dissociated,

7.1 x 10^-3 = x^2/0.03

x = [H3O+] = 0.0146 M

pH = -log[H3O+] = 1.84

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