Question

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solutions with the following pH. (a) -0.76 (b) 9.11 (c) 3.81 (d) 12.08

Short answer

Question: Calculate the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in the following solutions for each given pH value: (a) -0.76, (b) 9.11, (c) 3.81, and (d) 12.08. Answer: (a) For pH = -0.76 H+ concentration: \(\left[\mathrm{H}^{+}\right] = 10^{-(-0.76)} = 5.75 \times 10^{0}\) M pOH: \(14 - (-0.76) = 14.76\) OH- concentration: \(\left[\mathrm{OH}^{-}\right]=10^{-(14.76)} = 1.74 \times 10^{-15}\) M (b) For pH = 9.11 H+ concentration: \(\left[\mathrm{H}^{+}\right]=10^{-9.11} = 7.76 \times 10^{-10}\) M pOH: \(14 - 9.11 = 4.89\) OH- concentration: \(\left[\mathrm{OH}^{-}\right]=10^{-(4.89)} = 1.29 \times 10^{-5}\) M (c) For pH = 3.81 H+ concentration: \(\left[\mathrm{H}^{+}\right]=10^{-3.81} = 1.55 \times 10^{-4}\) M pOH: \(14 - 3.81 = 10.19\) OH- concentration: \(\left[\mathrm{OH}^{-}\right]=10^{-(10.19)} = 6.45 \times 10^{-11}\) M (d) For pH = 12.08 H+ concentration: \(\left[\mathrm{H}^{+}\right]=10^{-12.08} = 8.26 \times 10^{-13}\) M pOH: \(14 - 12.08 = 1.92\) OH- concentration: \(\left[\mathrm{OH}^{-}\right]=10^{-(1.92)} = 1.20 \times 10^{-2}\) M

Step-by-step solution

(a) Calculate H+ concentration from pH

To calculate \(\left[\mathrm{H}^{+}\right]\) in the solution with pH = -0.76, we use the formula: \(pH=-\log\left[\mathrm{H}^{+}\right]\). We will rearrange this formula to get \(\left[\mathrm{H}^{+}\right]=10^{-pH}\) and substitute the given pH value: \(\left[\mathrm{H}^{+}\right]=10^{-(-0.76)}\)

(a) Calculate pOH

Next, we need to calculate the pOH using the formula: \(pH + pOH = 14\). We will rearrange this formula to get \(pOH=14-pH\) and substitute the given pH value: \(pOH = 14 - (-0.76)\)

(a) Calculate OH- concentration from pOH

Now, we'll calculate \(\left[\mathrm{OH}^{-}\right]\) using the pOH value we just found. We'll use the formula: \(pOH=-\log\left[\mathrm{OH}^{-}\right]\) and rearrange it to get \(\left[\mathrm{OH}^{-}\right]=10^{-pOH}\). Substitute the pOH value: \(\left[\mathrm{OH}^{-}\right]=10^{-(14-(-0.76))}\) Repeat the same process for the remaining cases (b), (c), and (d).

(b) Calculate H+ concentration from pH

\(\left[\mathrm{H}^{+}\right]=10^{-9.11}\)

(b) Calculate pOH

\(pOH = 14 - 9.11\)

(b) Calculate OH- concentration from pOH

\(\left[\mathrm{OH}^{-}\right]=10^{-(14-9.11)}\)

(c) Calculate H+ concentration from pH

\(\left[\mathrm{H}^{+}\right]=10^{-3.81}\)

(c) Calculate pOH

\(pOH = 14 - 3.81\)

(c) Calculate OH- concentration from pOH

\(\left[\mathrm{OH}^{-}\right]=10^{-(14-3.81)}\)

(d) Calculate H+ concentration from pH

\(\left[\mathrm{H}^{+}\right]=10^{-12.08}\)

(d) Calculate pOH

\(pOH = 14 - 12.08\)

(d) Calculate OH- concentration from pOH

\(\left[\mathrm{OH}^{-}\right]=10^{-(14-12.08)}\)

Key concepts

H+ concentration

Understanding the concentration of hydrogen ions (\text{H}^{+}) in a solution is fundamental in chemistry, particularly in discussing acids and bases. The pH scale is a logarithmic scale used to represent the acidity or basicity of a solution. Mathematically, pH is the negative logarithm of the \text{H}^{+} concentration, expressed as:
\[ pH = -\text{log}[\text{H}^{+}] \].
By rearranging this formula, one can easily find the hydrogen ion concentration given the pH:
\[ [\text{H}^{+}] = 10^{-pH} \].
For instance, if the pH of a solution is -0.76, using the rearranged formula, the concentration of hydrogen ions would be calculated as \( [\text{H}^{+}] = 10^{-(-0.76)} \) which simplifies to \( [\text{H}^{+}] = 10^{0.76} \). This simple calculation can help students understand the direct relationship between pH and \text{H}^{+} concentration and is essential for comprehending acid-base balance.

pOH calculation

A complementary scale to pH is pOH, which represents the concentration of hydroxide ions (\text{OH}^{-}). It can be calculated using the relationship between pH and pOH, which together always add up to 14 at 25°C (298 K):
\[ pH + pOH = 14 \].
Let's illustrate this with an example. Given a solution with a pH of 9.11, to find its pOH, we simply subtract the pH from 14:
\[ pOH = 14 - pH \]\[ pOH = 14 - 9.11 = 4.89 \].
From this calculation, we can see how pH and pOH are interdependent, and understanding one allows for the determination of the other. A lower pH means a higher pOH, which indicates fewer available \text{OH}^{-} ions and thus a more acidic solution. Conversely, a higher pH indicates a lower pOH and a more basic solution.

OH- concentration

After calculating pOH as seen in the previous section, determining the hydroxide ion concentration (\text{OH}^{-}) follows a similar logarithmic relationship as pH:
\[ pOH = -\text{log}[\text{OH}^{-}] \].
To find the \text{OH}^{-} concentration, we use the inverse process:
\[ [\text{OH}^{-}] = 10^{-pOH} \].
For example, with a pOH of 4.89, the \text{OH}^{-} concentration would be:
\[ [\text{OH}^{-}] = 10^{-4.89} \].
This demonstrates that as the pOH decreases, the concentration of \text{OH}^{-} increases, indicating a more basic solution. The ability to convert between pH, pOH, and ion concentrations equips students with a powerful tool to understand the chemical nature of solutions, predict their behaviors, and address real-world problems such as water treatment and environmental monitoring.