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Once the correct mass of sodium acetate trihydrate has been weighed out, it will be quantitatively transferred to a 100
Once the correct mass of sodium acetate trihydrate has been weighed out, it will be quantitatively transferred to a 100.00 mL volumetric flask. Then, the pH of the buffer must be adjusted by adding an appropriate amount of strong acid. After the sodium acetate trihydrate has been completely transfered (with rinsing) into the volumetric flask with a small amount of deionized water, and the correct amount of HCl (calculated here) has been added, the buffer solution will be diluted to the final volume (100.00 mL).
How many moles of a strong monoprotic acid (hydrochloric acid) must be added to adjust the buffer pH to 4.45? You will use 0.500 mol/L hydrochloric acid for this task.
Expert Solution
Acid buffer solution is to be prepared . It will contain Acetic acid and sodium acetate Using Handerson equation ! PH = pka + logio [ CH 3 ( oona ] [CH3 (OOH) 4. 60 = 4. 74 + 10910 ( CH? (coNa] [ CH3 COOK ] Jogio [ CH 3 cooNa) = 4.60 - 4-74 ( CH 3(OOH] log10 [ CH3 coONa] = - 0.14 ( CH & COOH ] [ CH? (oONa) = 10-0.14 ( CHACOOH ) [CH3(OONa] = 0. 7244 O [ CH3 (OOH] Total buffer concentration is 0.125 moll L . Let 21 mos / L be the concentration of Acetic acid . The concentration of sodium accrate will be (0.125 - 2] mal/ L . Substitute values in ean O 0 . 125 - 21 = 0 . 7244 0 .125 - 21 = 0 . 7244 21 0 .125 = 21 + 0 . 724427 0 .125 = 1. 7244 2. 21 2 0.125 1. 7244 M2 0 . 0725
Hence the concentration of acetic acid in the buffer is 0 . 0725 mollL Total volume of buffer : 7210 oml So No. of Moles of Autic acid 0 . 0725 mol /L X loom! : 0. 00725 mol 1000mill When Hel is added to sodium acetate , acetic acid is obtained CHOCOONa + HCJ CHOCOOH+ Nace. Thus 0-00725 mol of Acetic acid in formed when 0 . 00725 moles of HC reacts with 0.00725 moles of sodium acetate . Hence 0.00725 mobs of HU must be added .
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