Suppose our initial concentration had been 0.10 for the acetic acid in the example problem we just M worked. For acetic acid, 100Ka = 1.8 × 10−3 or 0.0018. That’s essentially subtracting zero from 0.10 M. Aside from that if you did subtract it, you’d still follow the decimal place” subtraction sig. fig. rule “least and report 0.10 as your answer to the subtraction. M Example Calculate the pH of a 10.10 solution of acetic acid. The of acetic acid is 1.8 × 10−5. M Ka Reaction HC2H3O2 H+ C2H3O2− + ↔ Initial 0.10 0 0 Change +x +x −x Equilibrium 0.10 x x x − + − = = 1.8 ×10−5 K x2 2 3 2 = − 0.10 x HC2 H3O2 By cross multiplying and NOT neglecting you get = 1.8 × 10−6 1.8 × 10−5x ; collect like terms and x, x2 – use either the solver on your graphing calculator or a quadratic formula solving program you’ve loaded on your calculator (all of this to avoid arithmetic mistakes!) to solve for You should determine that x. = [H+] = 0.0013327 (way too many sig. figs!) x and that the pH = log (0.0013327) = 2.88 (2 SF) − Had we the math simplifies to x”, “neglected + − = = ×10−5 1.8 K x2 2 3 2 and = 1.8 × 10−6 x2 = 0.10 − x HC2 H3O2 So, take the square root of each side to get = 1.8 × 10−6 x2 = 0.00134 = x H+ pH = log (.001341) = 2.87 (2 SF) which is mighty, mighty close, so it is a really good approximation. − So, what’s the good news? The AP exam does not have equilibrium problems that require the quadratic formula.
Suppose our initial concentration had been 0.10 M for the acetic acid in the example problem we just worked. For acetic acid, 100Ka = 1.8 × 10−3 or 0.0018. That’s essentially subtracting zero from 0.10 M. Aside from that if you did subtract it, you’d still follow the “least decimal place” subtraction sig. fig. rule and report 0.10 M as your answer to the subtraction. Example Calculate the pH of a 10.10 M solution of acetic acid. The Ka of acetic acid is 1.8 × 10−5. Reaction HC2H3O2 ↔ H+ + C2H3O2− Initial Change Equilibrium 0.10 −x 0.10 − x 0 +x x 0 +x x + − K = 2 3 2 = 1.8 ×10−5 = HC2 H3O2 x2 0.10 − x By cross multiplying and NOT neglecting x, you get x2 = 1.8 × 10−6 – 1.8 × 10−5x ; collect like terms and use either the solver on your graphing calculator or a quadratic formula solving program you’ve loaded on your calculator (all of this to avoid arithmetic mistakes!) to solve for x. You should determine that x = [H+] = 0.0013327 (way too many sig. figs!) and that the pH = − log (0.0013327) = 2.88 (2 SF) Had we “neglected x”, the math simplifies to + − K = 2 3 2 = 1.8 ×10−5 = HC2 H3O2 x2 0.10 − x and x2 = 1.8 × 10−6 So, take the square root of each side to get x2 = 1.8 × 10−6 x = 0.00134 = H+ pH = − log (.001341) = 2.87 (2 SF) which is mighty, mighty close, so it is a really good approximation. So, what’s the good news? The AP exam does not have equilibrium problems that require the quadratic formula.
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