Question

600 mL of 0.01M HCl is mixed with 400 mL of 0.01 M H₂SO₄ .The pH of the mixture is ______ × 10^-2 (Nearest integer) 
[Given log 2 =0.30 , log 3 =0.48 log 5 =0.69  log 7 -0.84 log 11 =1.04 ]

Answer 1

Correct Answer: 186

To find the pH of the mixture formed by mixing 600 mL of 0.01M HCl with 400 mL of 0.01M H₂SO₄, we follow these steps:

Step 1: Calculate moles of H⁺ from HCl.
HCl dissociates completely in water:
\[\text{HCl} \rightarrow \text{H}⁺ + \text{Cl}^-\]
Initial moles of HCl = 0.01 moles/L × 0.6 L = 0.006 moles.
Moles of H⁺ from HCl = 0.006 moles.

Step 2: Calculate moles of H⁺ from H₂SO₄.
H₂SO₄ dissociates in two steps, but we'll assume complete dissociation for simplicity:
\[\text{H}_2\text{SO}_4 \rightarrow 2\text{H}⁺ + \text{SO}_4^2-\]
Initial moles of H₂SO₄ = 0.01 moles/L × 0.4 L = 0.004 moles.
Moles of H⁺ from H₂SO₄ = 2 × 0.004 = 0.008 moles.

Step 3: Total moles of H⁺ in the mixture.
Total H⁺ = 0.006 + 0.008 = 0.014 moles.

Step 4: Calculate concentration of H⁺.
Total volume of mixture = 600 mL + 400 mL = 1000 mL = 1 L.
\[\text{[H}⁺\text{]} = \frac{0.014 \text{ moles}}{1 \text{ L}} = 0.014 \text{ M}\]

Step 5: Calculate pH.
\[\text{pH} = -\log_{10}(0.014)\]
Using given logs:
\[-\log_{10}(0.014) = -(\log_{10}(14) - \log_{10}(10^3)) = 3 - \log_{10}(14)\]
\[\log_{10}(14) = \log_{10}(2 \times 7) = \log_{10}(2) + \log_{10}(7) = 0.30 + 0.84 = 1.14\]
\[\text{pH} = 3 - 1.14 = 1.86\]

Thus, the pH of the mixture is 186 × 10^-2, confirming it is within the range of 186,186.