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What would the pH of a solution containing 3.55g of an acid HX (Molar Mass of HX = 29.00 g/mol) after the addition of 2.90g o
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Answer #1

no of moles of HX = W/G.M.Wt

                            = 3.55/29   = 0.1224moles

no of moles of NaX = W/G.M.Wt

                             = 2.90/51   = 0.057moles

PKa   =-logKa

        = -log(2.14*10^-5)

        = 4.67

PH   = Pka + log[NaX]/[HX]

      = 4.67 + log(0.057/0.1224)

      = 4.67-0.3319

     = 4.3381 >>>>answer

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