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Exercises · 6.56

Q.The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

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The hydrogen ion concentration [H+][H^+] is found using [H+]=10−pH[H^+] = 10^{-\text{pH}}. For pH 6.8, 5.0, 4.2, 2.2, and 7.8, the corresponding [H+][H^+] values are 1.58×10−71.58 \times 10^{-7}, 1.0×10−51.0 \times 10^{-5}, 6.31×10−56.31 \times 10^{-5}, 6.31×10−36.31 \times 10^{-3}, and 1.58×10−81.58 \times 10^{-8} mol/L respectively.

The pH scale is a logarithmic measure of how acidic or basic a solution is. The key relationship is that pH is the negative logarithm (base 10) of the hydrogen ion concentration in moles per litre (mol/L). So, if you know the pH, you can always find [H+][H^+] by reversing that logarithm.

The formula is:

[H+]=10−pH[H^+] = 10^{-\text{pH}}

This means that a difference of 1 pH unit corresponds to a tenfold change in [H+][H^+]. A lower pH means a much higher [H+][H^+], and a higher pH means a much lower [H+][H^+]. Let's apply this to each substance.

  1. Milk (pH = 6.8)

[H+]=10−6.8[H^+] = 10^{-6.8}

To evaluate $10^{-6.8}$, it's helpful to rewrite it. $10^{-6.8} = 10^{-7 + 0.2} = 10^{-7} \times 10^{0.2}$.
We know $10^{0.3} \approx 2$, so $10^{0.2}$ is a bit less than 2. More precisely, $10^{0.2} \approx 1.58$.
Therefore, $[H^+] = 1.58 \times 10^{-7}$ mol/L.

2. Black Coffee (pH = 5.0)

[H+]=10−5.0=1.0×10−5 mol/L[H^+] = 10^{-5.0} = 1.0 \times 10^{-5} \text{ mol/L}

This is a clean, exact value because the pH is an integer.

3. Tomato Juice (pH = 4.2)

[H+]=10−4.2=10−5+0.8=10−5×100.8[H^+] = 10^{-4.2} = 10^{-5 + 0.8} = 10^{-5} \times 10^{0.8}

$10^{0.8}$ is a common value to remember. Since $10^{0.3010} \approx 2$, then $10^{0.8} = 10^{0.3010 \times 2.66} \approx 2^{2.66}$. A more direct approximation is $10^{0.8} \approx 6.31$.
So, $[H^+] = 6.31 \times 10^{-5}$ mol/L.

4. Lemon Juice (pH = 2.2)

[H+]=10−2.2=10−3+0.8=10−3×100.8[H^+] = 10^{-2.2} = 10^{-3 + 0.8} = 10^{-3} \times 10^{0.8}

Using the same $10^{0.8} \approx 6.31$ from above, we get:
$[H^+] = 6.31 \times 10^{-3}$ mol/L.

5. Egg White (pH = 7.8)

[H+]=10−7.8=10−8+0.2=10−8×100.2[H^+] = 10^{-7.8} = 10^{-8 + 0.2} = 10^{-8} \times 10^{0.2}

Using $10^{0.2} \approx 1.58$, we get: …
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