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Question

indow Help x(w ust of box?G How Man x(GABuffer sx alentActivity.do?locator assignment-take& ?CKe D Specific H ×( Desmos l Table o, A vs PKa data x the References to access important values if needed for this auestiaa. April 14, 2018, 6:04 PM Determine the pH change when 0.006 mo, HCI is added to 1.0° L of a buffer solution that is 0.459 M in HCN and 0.390 M in CN. pH after addition- pHi before addition pH change Determine the pH change when 0.134 moi KOH is added to 1.00 L of a buffer solution that is 0.460 M in HNO2 and 0.317 M in NO pH after addition-pH before addition pH change- 4) 7 8. 9

Explanation / Answer

Answer – We are given, [HCN] = 0.459 M

[CN-] = 0.390 M , volume = 1.0 L

First we need to calculate pH before added any acid

We know the pKa value for the HCN and it is 9.21

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [CN-] / [HCN]

      = 9.21 + log 0.390 / 0.459

      = 9.14

pH after added 0.086 moles of HCl

Moles of HCN = 0.459 M x 1.0 L

                        = 0.459 moles

Moles of CN- = 0.390 M x 1.0 L

                       = 0.390 moles

when we added HCl the moles of acid increased, and its conjugate base decreased

moles of HCN = 0.459 moles + 0.086 mole = 0.545 moles

moles of CN- = 0.390 moles – 0.086 moles = 0.304 moles

total volume = 1.0 L

so, [HCN] = 0.545 moles / 1.0 L = 0.545 M

[CH3COO-] = 0.304 mole / 1.0 L = 0.304 M

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [CN-] / [HCN]

      = 9.21 + log 0.304 / 0.545

      = 8.95

Change in pH = 9.14 – 8.95

                       = 0.183

b) We are given, [HNO2] = 0.460 M

[NO2-] = 0.317 M , volume = 1.0 L

First we need to calculate pH before added any base

We know the pKa value for the HNO2 and it is 3.39

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [NO2-] / [HNO2]

      = 3.39 + log 0.317 / 0.460

      = 3.23

pH after added 0.134 moles of KOH

Moles of HNO2 = 0.460 M x 1.0 L

                          = 0.460 moles

Moles of NO2- = 0.317 M x 1.0 L

                       = 0.317 moles

when we added KOH the moles of acid decreased, and its conjugate base increased

moles of HNO2 = 0.460 moles -0.134 mole = 0.330 moles

moles of NO2- = 0.317 moles + 0.134 moles = 0.451 moles

total volume = 1.0 L

so, [HNO2] = 0.330 moles / 1.0 L = 0.330 M

[NO2-] = 0.451 mole / 1.0 L = 0.451 M

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [NO2-] / [HNO2]

      = 3.39 + log 0.451 / 0.330

      = 3.53

Change in pH = 3.53 – 3.23

                       = 0.297

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