Air & noise pollution
Quarry waste
Damage to biodiversity
Answer:
Explanation:
No not always but it is more likely they will
a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
<em />
Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
