Calculate the pH of 100.0 mL of a 0.150M aqueous solution of HNO3

1 answer:

7 0

PH scale is used to determine how acidic, neutral or basic a solution is.
pH can be calculated using the following equation
pH = -log[H⁺]
HNO₃ is a strong acid therefore completely dissociates into its ions as follows
HNO₃ --> H⁺ + NO₃⁻
1 mol of HNO₃ dissociates to produce 1 mol of H⁺ ions
Therefore [HNO₃] = [H⁺]
[H⁺] = 0.150 M
since concentration has been given, its not necessary to consider the volume as concentration is constant throughout the whole solution
pH = -log(0.150 M)
pH = 0.82
the pH of solution is 0.82

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L

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Answer:

$\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}$

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

$P_1V_1=P_2V_2$

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

$25.0 \ L * 2.05 \ atm = P_2V_2$

The volume is decreased to 14.5 liters, but the pressure is unknown.

$25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L$

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}$