Question 10 of 15 eMA Sapling Learning this question hasbeamantmaad Map A Rendam

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Question 10 of 15 eMA Sapling Learning this question hasbeamantmaad Map A Rendamhamcontriat Lindenweoduniversity Two aqueous solutions are prepared. The first is a 1.ox102 mL solution of has a 1.75. The second is a 1.150 L solution of HBr that has a pH of 4.50. When these two salutions mixed what are is the pH, Hao' and loH1 of he resulting solution? pH poH a Number Number [OH1 o Previous ® Give up 8 verw solution o check Answer 0 Next HExit creers partners Privacy policy terms of use contact us help

Explanation / Answer

Consider the dissociation of HClO4 and HBr as below:

HClO4 <=====> H+ + ClO4-

HBr <=====> H+ + Br-

Both the acids are monoprotic acids and hence 1 mole of each acid must furnish 1 mole of proton. Number of moles are additive and hence we aim to find out the total number of moles of protons when the two acids are mixed.

We define pH of a solution as pH = -log [H+]; therefore,

[H+] = antilog (-pH) = 10^(-pH)

Use the pH of the two acids to compute the [H+] for each:

HClO4: [H+] = 10^(-1.75) = 0.01778 M

HBr: [H+] = 10^(-4.50) = 3.16228*10-5 M

Next compute the mole(s) of H+ contributed by each acid by multiplying the molar concentration of proton by the volume of the acid in litres.

HClO4: (1.0*102 mL)*(1 L/1000 mL)*(0.01778 mol/L) = 0.001778 mol.

HBr: (1.50 L)*(3.16228*10-5 mol/L) = 4.74342*10-5 mol.

Compute the total number of moles of proton by adding the two.

Total moles of proton = (0.001778 + 4.74342*10-5) mole = 1.8254342*10-3 mole.

Compute the molar concentration of proton in the mixed solution as below.

Total volume of solution = (1.0*102 mL)*(1 L/1000 mL) + (1.50 L) = 1.60 L

Molar concentration of H+ = (1.8254342*10-3 mole)/(1.60 L) = 1.14089*10-3 mol/L 0.001141 mol/L 0.00114 M.

Compute the pH, pOH and [OH-] of the solution as below:

pH = -log [H+] = -log (0.00114) = 2.943 2.94

pOH = 14 – pH = 14 – 2.94 = 11.06

We define pOH as pOH =-log [OH-] so that [OH-] = 10^(-pOH) = 10^(-11.06) = 8.7096*10-12 M 8.71*10-12 M.

We know that a bare proton, H+ cannot exist in an aqueous solution and always attaches to a proton; hence, [H+] = [H3O+].

Ans:

pH = 2.94

pOH = 11.06

[H3O+] = 0.00114 M

[OH-] = 8.71*10-12 M

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