CHEM 1322 Lecture Notes - Lecture 10: Rice Chart, Equivalence Point
11 Mar 2017
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150ml of 0. 20m acetic acid, ka = 1. 7x10-5 titrated with 0. 30m sodium hydroxide (strong base) 125: at start, we only have 150ml of 0. 20m acetic acid (our weak acid). To find ph, we know [h+] = (cid:1837) = (cid:883). 7(cid:883)(cid:882)^-5 x (cid:882). (cid:884)(cid:882)m = . 00184390. Take -log10(. 00184390) to get: next, when we start titrating, we have 0. 030 mol of weak acid (0. 20m x . 150l) and ph = 2. 73. We know x = 0. 0075 because all of the base reacts. So, we can plug into the ka equation. ***remember: divide by total volume and use concentrations in the equation, even though we used moles in the table. [h+] = 5. 11x10-5 ph = 4. 29: for the half equivalence point, [h+] = ka, so ph = -log(1. 7x10-5) = 4. 77, make a table again now, similar to step 2. [h+] = 5. 67x10-6 ph = 5. 25: start the next step by finding the amount of product. We know that kb = kw/ka = 1. 0x10-14/1. 7x10-5.
