Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Answer:
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“Inquisitive” Explain why your subject is important, curious, or interesting.
“Paradoxical” Explain what aspects of your subject seem improbable. ...
“Corrective” Explain how your subject has been misunderstood or misrepresented by others. ...
Answer:
1.3 × 10³ mL
Explanation:
Let's consider the following reaction.
Zn + 2 HCl → ZnCl₂ + H₂
The percent yield is 78.0%. The real yield (R) of zinc chloride is 35.5 g. The theoretical yield (T) of zinc chloride is:
35.5 g (R) × (100 g T/ 78.0 g R) = 45.5 g T
The molar mass of zinc chloride is 136.29 g/mol. The moles corresponding to 45.5 g of zinc chloride is:
45.5 g × (1 mol/ 136.29 g) = 0.334 mol
The molar ratio of HCl to ZnCl₂ is 2:1. The moles of HCl that react with 0.334 moles of ZnCl₂ are 2 × 0.334 mol = 0.668 mol.
We need 0.668 moles of a 0.50 M HCl solution. The volume required is:
0.668 mol × (1000 mL/0.50 mol) = 1.3 × 10³ mL
