PracticeSheet 15 key

PracticeSheet 15 key - Practice Sheet 15 BLC & DLC Answers...

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Practice Sheet 15 15B.1 For these problems, we will need to use the formula pH = –log[H + ]. a) [H + ] = 10 –pH = 10 –4.73 = 1.86 × 10 –5 M b) pH = –log[H + ] = –log (1.57 × 10 –3 ) = 2.80 c) [H + ] = 10 –1.34 = 0.0457 M d) pH = –log(0.95) = 0.022 e) pH = –log(0.01) = 2 f) [H+] = 10 –pH = 10 -10 = 1 × 10 -10 M 15B.2 Two relationships may be used here: pH + pOH = 14.00, and [H + ][OH ] = K w . We will use the first relationship to find the pOH, and then use 10 - pOH to find [OH - ]. a) pOH = 14.00 – pH = 14.00 – 4.73 = 9.27 [OH ] = 10 –9.27 = 5.37 × 10 –10 M b) pOH = 14.00 – 2.80 = 11.20 [OH ] = 10 –11.20 = 6.310 × 10 –12 M c) pOH = 14.00 – 1.34 = 12.66 [OH ] = 10 –12.66 = 2.188 × 10 –13 M d) pOH = 14.000 – 0.022 = 13.978 [OH ] = 10 –13.978 = 1.0520 × 10 –14 M e) pOH = 14 – 2 = 12 [OH ] = 10 –12 = 1 × 10 –12 M f) pOH = 14 – 10 = 4 [OH ] = 10 –4 = 1 × 10 –4 M 15B.3 2H 3 PO 4 + 3Ca(OH) 2 → Ca 3 (PO 4 ) 2 + 6H 2 0 Using this balanced equation:
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This note was uploaded on 11/04/2011 for the course CHEM 105 taught by Professor Macedone during the Fall '07 term at BYU.

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PracticeSheet 15 key - Practice Sheet 15 BLC & DLC Answers...

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