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re from Cengage Learning - Google Chrome ilrn/takeAssignment/takeCovalentActivity.do?locator-assignmen Review Topics] [References] Use the References to access important values if needed for this question. A buffer solution contains 0.309 M ammonium bromide and 0.258 M ammonia. If 0.0211 moles of sodium hydroxide are added to 125 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume change does not change upon adding sodium hydroxide) Submit Answer Retry Entire Group 7 more group attempts remaining

Explanation / Answer

No. Of moles of ammonia = molarity × vol in litre = 0.258 × 0.125L = .03225 moles .

No. Of moles of ammonium bromide = 0.309 × 0.125 = 0.038625 moles .

No. Of moles of NaOH = 0.0211 moles .

No. Of moles of ammonia (base) will increase after the addition of NaOH and no. Of moles of ammonium bromide (salt) will decrease after the addition of NaOH .

So the , pOH = pKb + log [ salt /base ]

pKb for ammonia is 4.74

pOH = 4.74 + log ( 0.038625 - 0.0211 / 0.03225+0.0211)

pOH = 4.74 + log ( 0.017525 - 0.05335)

pOH = 4.74 - 0.48 = 4.26

pH = 14 - pOH = 14 - 4.26 = 9.74

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