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Calculating pH of solutions?

Calculating pH for biology lab?

  • Today in my bio lab I measured the pH of lemon juice, coffee, and sea water with pH paper and a pH meter. The next part of the lab wants me to " Calculate the following values using the pH values obtained with pH meter for pOH, H3O and OH", only I have no idea how to do that. These are the values I got for the pH paper: coffee- 3.5 Lemon juice- 2.5 Sea water- 7.5 And heres the values with the pH meter: Coffee- 5.17 Lemon juice- 2.34 Sea Water- 7.04 Any help figuring this out is appreciated!

  • Answer:

    The pOH of a solution is the concentration of OH⁻ ions. If you imagine pH and pOH on a number line, the middle is 7 with everything to the left on the number line (0 to <7) being acidic and everything to the right on the number line (>7 to 14) being basic. A highly acidic compound has a low pH (1-2) and a high pOH (13-14), and a highly basic compound has a high pH (13-14) and a low pOH (1-2). The relationship of pH and pOH is that they are on equally opposite sides of the number line. A solution with a pH of 5 would have a pOH of 9. To get the pOH, you use the equation pOH = 14 - pH. Coffee (pH = 5.17) has a pOH of 14 - 5.17 = 8.83 Lemon Juice (pH = 2.34) has a pOH of 14 - 2.34 = 11.66 Sea Water (pH = 7.04) has a pOH of 14 - 7.04 = 6.96 For H₃O⁺ and OH⁻, I am assuming it means to calculate the concentration of those ions within the three solutions. So, to calculate the concentration of H₃O⁺ in solution you use the equation pH = -log[H⁺]. And, for the concentration of OH⁻ in solution you use the equation pOH = -log[OH⁻]. Coffee (pH = 5.17) => 5.17 = -log[H₃O⁺] => [H₃O⁺] = 6.76 * 10^(-6) mol/L Coffee (pOH = 8.83) => 8.83 = -log[OH⁻] = [OH⁻] = 1.48 * 10^(-9) mol/L Lemon Juice (pH = 2.34) => 2.34 = -log[H₃O⁺] => [H₃O⁺] = 4.57 * 10^(-3) mol/L Lemon Juice (pOH = 11.66) => 11.66 = -log[OH⁻] = [OH⁻] = 2.19 * 10^(-12) mol/L Sea Water (pH = 7.04) => 7.04 = -log[H₃O⁺] => [H₃O⁺] = 9.12 * 10^(-8) mol/L Sea Water (pOH = 6.96) => 6.96 = -log[OH⁻] = [OH⁻] = 1.10 * 10^(-7) mol/L

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The pOH of a solution is the concentration of OH⁻ ions. If you imagine pH and pOH on a number line, the middle is 7 with everything to the left on the number line (0 to <7) being acidic and everything to the right on the number line (>7 to 14) being basic. A highly acidic compound has a low pH (1-2) and a high pOH (13-14), and a highly basic compound has a high pH (13-14) and a low pOH (1-2). The relationship of pH and pOH is that they are on equally opposite sides of the number line. A solution with a pH of 5 would have a pOH of 9. To get the pOH, you use the equation pOH = 14 - pH. Coffee (pH = 5.17) has a pOH of 14 - 5.17 = 8.83 Lemon Juice (pH = 2.34) has a pOH of 14 - 2.34 = 11.66 Sea Water (pH = 7.04) has a pOH of 14 - 7.04 = 6.96 For H₃O⁺ and OH⁻, I am assuming it means to calculate the concentration of those ions within the three solutions. So, to calculate the concentration of H₃O⁺ in solution you use the equation pH = -log[H⁺]. And, for the concentration of OH⁻ in solution you use the equation pOH = -log[OH⁻]. Coffee (pH = 5.17) => 5.17 = -log[H₃O⁺] => [H₃O⁺] = 6.76 * 10^(-6) mol/L Coffee (pOH = 8.83) => 8.83 = -log[OH⁻] = [OH⁻] = 1.48 * 10^(-9) mol/L Lemon Juice (pH = 2.34) => 2.34 = -log[H₃O⁺] => [H₃O⁺] = 4.57 * 10^(-3) mol/L Lemon Juice (pOH = 11.66) => 11.66 = -log[OH⁻] = [OH⁻] = 2.19 * 10^(-12) mol/L Sea Water (pH = 7.04) => 7.04 = -log[H₃O⁺] => [H₃O⁺] = 9.12 * 10^(-8) mol/L Sea Water (pOH = 6.96) => 6.96 = -log[OH⁻] = [OH⁻] = 1.10 * 10^(-7) mol/L

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