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.

The key idea is that pH is the negative logarithm (base 10) of the hydrogen ion concentration:

$[H^+] = 10^{-\text{pH}}$.

Step 1: For each substance, substitute the given pH into $[H^+] = 10^{-\text{pH}}$.

Step 2: Compute the value. For example, milk: $[H^+] = 10^{-6.8} = 10^{-7} \times 10^{0.2} \approx 1.58 \times 10^{-7} \, \text{M}$.

Step 3: Repeat for the others.

• Milk (pH 6.8): $[H^+] = 10^{-6.8} \approx 1.58 \times 10^{-7} \, \text{M}$
• Black coffee (pH 5.0): $[H^+] = 10^{-5.0} = 1.0 \times 10^{-5} \, \text{M}$
• Tomato juice (pH 4.2): $[H^+] = 10^{-4.2} \approx 6.31 \times 10^{-5} \, \text{M}$
• Lemon juice (pH 2.2): $[H^+] = 10^{-2.2} \approx 6.31 \times 10^{-3} \, \text{M}$ …

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 $[H^+]$ by reversing that logarithm.

The formula is:

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

1. Milk (pH = 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)

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

3. Tomato Juice (pH = 4.2)

$10^{0.8}$ 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)

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)

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