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In: Chemistry

a.) If 5 ml of 0.1 NaOH (a strong base) is added to 100 ml of...

a.) If 5 ml of 0.1 NaOH (a strong base) is added to 100 ml of 0.05 phosphate buffer (phosphoric acid- H3PO4), pH 7.1, what is the resulting pH? (Ignore the volume change). Clearly identify the appropriate acid and conjugate base and determine their concentration in the final solution. b.) What would the resulting pH be if instead you added 5 ml of 0.25 M HCl? (Ignore the volume change) *Please show your work and steps to understand*

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Expert Solution

pH = pka + log [ HPO42-] /[H2PO4- ]    since pH is near pka2 , we take H2PO4- and HPO42- buffer

7.1 = 7.2 + log [ HPO42-] /[H2PO4-]

[H2PO4-] = 0.7943 [HPO42-] ............(1)

given Buffer conc = 0.05 = [H2PO4-] + [HPO42-] ..........(2)

solving ( 1) ( 2) we get [H2PO4-] = 0.02213   , [HPO42-] = 0.02787

H2PO4- moles = M x V = 0.02213 x 0.1 = 0.002213 , HPO42- moles = 0.002787

NaOH moles added = M x V = 0.1 x 5/1000 = 0.0005

after reacting with NaOH , H2PO4- moles = 0.002213-0.0005 = 0.001713 ( base reacts with acid )

HPO42- moles = 0.002787+0.0005 = 0.003287

pH = 7.2 + log ( 0.003287/0.001713) = 7.5    ( vol term cancells out in numerator and denominator in concentration )

now if HCl added, HCl moles = M x V = 0.25 x 5/1000 = 0.00125

H2PO4- moles = 0.002213+0.00125 = 0.003463

HPO42- moles = 0.002787 -0.00125= 0.001537

pH = 7.2 + log ( 0.001537/0.003463) = 6.85

queen_honey_blossom answered 2 years ago
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