toru smo xie My Drive-Google Drive ?Chapter 18-principles c × X Cengage x y owlv

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toru smo xie My Drive-Google Drive ?Chapter 18-principles c × X Cengage x y owlv2] Online teachin now.com/irn/takeAssignment/takeCovalentActivitydo?iocator rassignment-take&takeAssignmentSession; Use the References to access important valoes id A butter solubon cootains 0.461 M NH,Br and 0.278 M NH, (ammonía). Determne the pH change when 0.077 mol NaOH is added to 1.00L of the buffer pH after addition-pH before addition pH change your answer Check for tys miscalculations efo before submting your answer Retry Entre Group 1 more group atternpt remaining

Explanation / Answer

We know that for a buffer system,

pOH = pKb + log { [salt] / [base] }

Ammonia (NH3) is a weak base with a pKb of 4.75.

So,

pOH = 4.75 + log (0.461 / 0.278)

pOH = 4.75 + log (1.66)

pOH = 4.75 + 0.22

pOH = 4.97

So, pH = 14 – pOH

            = 14 – 4.97

           = 9.03

Since you have 1.00 L, the moles of each form in the original solution are

Moles of NH4Br = 0.461 moles
Moles of NH3 = 0.278 moles

When 0.077 mol of NaOH is added, it will react with 0.077 mol of NH4Br to produce 0.077 mol of NH3.

So, moles after addition of NaOH

Moles of NH4Br = 0.461 moles – 0.077 moles = 0.384 moles
Moles of NH3 = 0.278 moles + 0.077 moles = 0.355 moles

Volume is 1.00 L

So, molarity terms are

[NH4Br] = 0.384 M
[NH3] = 0.355 M

So,

pOH = 4.75 + log (0.384 / 0.355)

pOH = 4.75 + log (1.08)

pOH = 4.75 + 0.03

pOH = 4.78

So, pH = 14 – pOH

            = 14 – 4.78

           = 9.22

Now,

pH change = pH after addition – pH before addition

                 = 9.22 - 9.03

                 = 0.19

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