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A buffer is prepared by dissolving 0.80 moles of NH3 and 0.80 moles of NH4Cl in 1.00 L of aqueous solution. If 0.10 mol of NaOH is added to 250 mL of this buffer, what is the pH of the resultant solution?
a.10.54
b.4.27
c.8.78
d.9.73
e.5.22

Respuesta :

baraltoa

Answer:

a. 10.54

Explanation:

reaction is

NH₃ +  NaOH   -----------------NH₄Cl + H₂O

0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl

mol NH₃  = 0.80 - 010 = .70

mol NH₄Cl = 0.80 + .10 = 0.90

pH = pKₐ + log (( NH₃/NH₄Cl))

pH =  9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54

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