Call ascorbic acid H2A to save typing. First determine the molarity of the NaOH from pH = 14.436.
pH = 14.436 so pOH = -0.436.
Use pOH = -log(OH^-) and I get OH^- = 2.7 but you need to get a more accurate number and redo the rest of the problem since my work is just an estimate.
......H2A + 2NaOH ==> Na2A + 2H2O
millimols NaOH = mL x M = 54.47 x 2.72 = approx 148
mmols H2A = 35.7 x 2.01 = approx 72
Convert mmols H2A to NaOH by 72 x 2 = approx 142.
148 = mmols NaOH initially.
143 = mmols NaOH used.
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5 mmols NaOH left.
M NaOH = mmols/mL = approx 6/90 = approx 0.05. Convert that to pOH then to pH. I get approx 1.5 for pH. Post your work if you get stuck.
What will the pH of a solution be when 0.5447 L of NaOH (pH = 14.436) is added to
0.357 L of 2.01 M ascorbic acid (H2C6H6O6)? Not sure how to do this.
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