1)Calculate the hydronium ion concentration and the pH of the solution that resu

Question

1)Calculate the hydronium ion concentration and the pH of the solution that results when 10.6 mL of 0.21 M acetic acid, is mixed with 6.9 mL of 0.060 M NaOH.

Hydronium ion concentration =?M

pH =

2)A buffer solution with a pH of 12.00 consists of Na3PO4 and Na2HPO4. The volume of solution is 200.0 mL. If the concentration of Na3PO4 is 0.300 M, what mass of Na2HPO4 is present?

2B) A buffer solution with a pH of 12.00 consists of Na3PO4 and Na2HPO4. The volume of solution is 200.0 mL. The concentration of Na3PO4 is 0.300 M.What mass of Na3PO4 is required to change the pH to 12.25?Ka = 3.6 × 10-13

Explanation / Answer

1) Initial acetic acid moles = M x V = 0.21 x ( 10.6/1000) = 0.002226

NaOH moles = 0.06 x 6.9/1000 =0.000414

acetic acid + NaOH ---> sodium actetate + H2O

hence acteic acid moles after reactiong with NaoH = 0.002226-0.000414 = 0.001812

sodium acetate moles = 0.000414

totla vol = 10.6+6.9 = 17.5 ml = 0.0175 L

pH = pka + log [sodium acetate ] / [acetic acid]

pH = 4.745 + log ( 0.000414/0.0175) / ( 0.001812/0.0175)

= 4.1

2) pH = pka of HPO42-   + log [Na3PO4]/[Na2HPO4]

12 = 12.4437 + log (0.3) /[Na2HPO4]

[Na2HPO4] = 0.833

moles of Na2HPO4 M x V = 0.8333 x ( 200/1000) = 0.1667

Na2HPO4 mass = moles x molar mass of Na2HPO4 = 0.1667 x 141.96 = 23.66 g

B)   pH = pka + log [Na3PO4]/[NaH2PO4]

12.55 = 12.4437 + log [Na3PO4]/[NaH2PO4]    ( pka = -log Ka = -log ( 3.6x10^-13) = 12.4437)

[Na3PO4] = 0.783 [Na2HPO4]

Na3PO4 moles = 0.783 Na2HPO4 moles , let m be moles of Na3PO4 added

then initialy Na3PO4 moles = M x V = 0.3 x 0.2 = 0.06 , after some Na3PO4 addition moles = 0.06+m

hence 0.06 + m = 0.783 ( 0.1667)

m = 0.0705

Na3PO4 mass needed = moles x molar mas s= 0.0705 x 163.94 = 11.56 g

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