03 12/8/2016 11:00 PM N608h00 12/8/20160801 PM Gradebook t Print calcul aloe Per

Question

03 12/8/2016 11:00 PM N608h00 12/8/20160801 PM Gradebook t Print calcul aloe Periodic Table Question 2 of 13 Incarr Sapling Learning Map 500.0 mL of 0.130 M NaoH is added to 535 mL of0250 M weak acid (K. 5.74 x104). What is the pH of the resulting buffer? HA(aq)+ oH (ag) ,00+ A (a4) Number pH 4.63 O Previous Give up & View solution check Answer Hint start by finding the number of moles (or milimoles) of HA and oH Which reactant is limiting? Next, find the amount of A that forms and the amount of HA left over, the LATMHA) ratio and use the Henderson-Hasselbalch equation lo fnd pH

Explanation / Answer

1) [NaOH] = molarity x volume in Litres = 0.13 M x 0.5 L = 0.065 mol

[weak acid] = [HA] = 0.25 M x 0.535 L = 0.13375 mol

Ka = 5.74x 10-5

HA + NaOH -----------------------> NaA + H2O

0.13375 mol 0.065 mol 0

---------------------------------------------------------------------------------------

0.13375-0.065 0 0.065 mol

= 0.06875 mol

Hence,

[HA] = 0.06875 mol

[NaA] = 0.065 mol

Then,

Henderson-Hasselbalch equation is

pH = - logKa + log [NaA]/[HA]

= - ( 5.74x 10-5) + log (0.065 mol/0.06875 mol )

= 4.2

pH = 4.2

Therefore,

pH of the resulting buffer = 4.2

2) Given that

[ weak acid] = 0.18 M

[conjugate base]= 0.4 M

Ka = 2.6 x 10-5

Acc to Henderson-Hasselbalch equation

pH = - logKa + log [conjugate base]/[weak acid]

= - ( 2.6 x 10-5) + log (0.4 M /0.18 M )

= 4.93

pH = 4.93

Therefore,

pH of the buffer = 4.93

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