How would you prepare 500.00 mL of a buffer with a pH of 6.00 from 0.5000 M Na_3
ID: 959497 • Letter: H
Question
How would you prepare 500.00 mL of a buffer with a pH of 6.00 from 0.5000 M Na_3 AsO_4 and 0.400 MHCI? identify the principal conjugate acid/base pair and calculate the ratio between them in a solution that is buffered to pH 7.50 and was originally made with 0.15 M H_3 AsO_4. How many grams of dipotassium phthalate (C_6 H_4(COOK)_2) must be added to 750 mL of 0.0500 M phthalic acid (C_6 H_4 (COOH)_2) to give a buffer of pH 5.75? For phthalic acid. pK_1 = 2.95 and pK_2 = 5.41 for phthalic acid. Tris(hydroxymethyl) ammomethane (Tris base) is used with Tris hydrochloride (Tris HCI) as an important physiological buffer. It is one of the most common biochemical buffers with uses in electrophoresis, treatment of cardiac arrest, to increase the solubility of cell membranes and others. Calculate the pH of a solution prepared by dissolving 10.0 g of Tris base (H_2 NC(CH_2 OH)_3), plus 10.0 g of tris hydrochloride (H_2 NC(CH_2 OH)3) in 0.250 L of water. (K_a = 8.32 times 10^9 for Tris HCI) What will the pH be if 10.5 mL of 0.500 M NaOH is added? To a 100.0 mL volumetric flask were added 7.00 mL of 1.70 M of H_3 VO_4, 1.401 g of NaH_2 VO_4 and 15.00 mL of 1.5 M NaOH. The flask was filled to the calibration mark with water. What were the molar concentrations of H^+ OH^-, the two other relevant molecules and the pH of the solution? Is the resulting solution a buffer? K_a1 = 1.0 times 10^-4, K_a2 = 2.8 times 10^-9 and K_a3 = 5.0 times 10^-15 for H_3 VO_4.Explanation / Answer
12.(a): moles of tris base added = mass / molar mass = 10.0 g / 121.14 g/mol = 0.08255 mol
moles of tris HCl added = mass / molar mass = 10.0 g / 157.6 g/mol = 0.06345 mol
Ka = 8.32x10-9
=> pKa = - log(8.32x10-9 ) = 8.08
pH of tris base/trisHCl buffer can be calculated from Hendersen equation.
pH = pKa + log[tris base] / [tris HCl]
=> pH = pKa + log (moles of tris base) / (moles of trisHCl) [ Since volume remains same for both]
=> pH = 8.08 + log(0.08255 mol / 0.06345 mol) = 8.19 (answer)
(b): Moles of NaOH added = MxV = 0.500 mol/L x 0.0105 L = 0.00525 mol
0.00525 mol NaOH will react with 0.00525 mol TrisHCl to form 0.00525 mol of Tris base.
Hence final moles of TrisHCl = 0.06345 mol - 0.00525 mol = 0.0582 mol
Final mol of Tris base = 0.08255 mol + 0.00525 mol = 0.0878 mol
pH = pKa + log[tris base] / [tris HCl]
=> pH = pKa + log (moles of tris base) / (moles of trisHCl) [ Since volume remains same for both]
=> pH = 8.08 + log(0.0878 mol / 0.0582 mol) = 8.26 (answer)
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