C. pH of Buffer Solutions In the unshaded portions of the following supply the c

Question

C. pH of Buffer Solutions In the unshaded portions of the following supply the calculated molarities of the indicated species in the table, Assume the solutions are at 25 C. solution mixturcs. Note that in the rows labeled "Before Reacti you must supply the appropri- ate concentrations after dilution occurs in the reaction mixture, but before any reaction takes place. Calculated Molarities in Mixtures [HCI] [NaOH) C.1 Buffer C2a Buffer HCI Before Reaction O.u8M 0.4q M 0,0STM At Equilibrium 0,5 C2b Buffer NaOH Before Reaction 6 M 0 .Yam At Equilibrium 0,4llM 0.S4 qM 1.30 C.3 Water At Equilibrium C.4a Water HCI Before Reaction At Equilibrium 0,0 SAm .4b Water NaOH Before Reaction At Equilibrium

Explanation / Answer

4a) Consider the dissociation of aqueous HCl as below:

HCl (aq) -------> H+ (aq) + Cl- (aq)

Consider the 1:1 nature of dissociation. As per the stoichiometric equation,

1 mole HCl = 1 mole H+

Therefore, 0.059 M HCl = 0.059 M H+ (ans)

We consider only [H+] since H+ is the species responsible for controlling pH of the solution. Moreover, HCl is a strong acid and dissociates completely into H+

pH = -log [H+] = -log (0.059) = 1.229 1.23 (ans).

4b) Consider the dissociation of aqueous NaOH as below:

NaOH (aq) ------> Na+ (aq) + OH- (aq)

Again, the electrolyte dissociates in a 1:1 manner. As per the stoichiometric reaction,

1 mole NaOH = 1 mole OH-

Therefore, [OH-] = 0.059 M. The above is true since NaOH is a strong electrolyte and dissociates completely into OH-.

Calculate the pOH = -log [OH-] = -log (0.059) = 1.229 1.23.

We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14 – 1.23 = 12.77.

Again, pH = -log [H+]; therefore,

[H+] = antilog (-pH) = antilog (-12.77) = 1.698*10-13 M 1.70*10-13 M (ans).

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