University of Kentucky-CHE1X Chom21Labs m/ibiscms/mod/ibis/view.php?id-4735338 4

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University of Kentucky-CHE1X Chom21Labs m/ibiscms/mod/ibis/view.php?id-4735338 4 Jump cky-CHE 113 Lab-Spring18-FRENCH Activities and Due Dates Pre-Lab 16 3/28/2018 05:00 PM 4 242/25 Gradebook Print CalcuistorPeriodic Table O formic acid, K-1.77x10 2.00 M O acetic acid, K, 1.75x10, 5.00 M Scroll down for more. sodium disulfate monohydrate, Kp1.20x10"2, 3.00 M O ? phosphoric acid, K.-752x 10", 100 M Conjugate bases O sodium suilfate decahydrate, NaSO 10H-o O sodium formate, HCOONa sodium acetate trihydrate, down to view more The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M Based on this information, what mass of solid conjugate base should the student make the buffer solution with a pH-2.5 Based on this information, what volume of acid should the student measure to make out to the 0.100 M buffer solution? Number Number 3.28 mL °Previous ®Givu Up & View Solton #Try Agan ONed fl Ex8 about us carers privacy policy serms of ue contacts

Explanation / Answer

Since you have answered most of the question, I shall take off from where you left. The Ka of phosphoric acid is 7.52*10-3; therefore, pKa = -log Ka = -log (7.52*10-3) = 2.12.

The pH of the desired buffer is 2.12. It is given that the concentration of the weak acid in the buffer solution must be 0.100 M. The final volume of the solution is 100.00 mL. The supplied stock of H3PO4 is 1.00 M.

Use the dilution equation as

C1*V1 = C2*V2

where C1 = 1.00 M; C2 = 0.100 M and V2 = 100.00 mL.

Plug in values and get

(1.00 M)*V1 = (0.100 M)*(100.00 mL)

====> V1 = (0.100 M)*(100.00 mL)/(1.00 M) = 10.00 mL.

We need to take 10.00 mL of 1.00 M supplied H3PO4 of 1.00 M, add 3.28 g of sodium dihydrogen phosphate monohydrate in a 100.00 mL volumetric flaks and dilute up to the mark with deionized water.

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