2- A 0.310 M solution of a weak acid, HX, has a pH of 2.53
a. Find the [H+] and the percent ionization of nitrous acid in this solution.
b. Write the equilibrium expression and calculate the value of Ka for the weak acid.
c. Calculate the pH of the solution formed by adding 10 g of KX (molar mass = 59.0 g•mol–1) to 650 mL of 0.0125 M solution of the weak acid.
d. Determine the pH of a solution formed by adding 10 g of KX to 750 mL of water.
I'm confused by the question. Is HX in the preamble the same as HNO2 is part a or is that different?
What is the molarity of a solution prepared by dissolving 4.92 g of calcium nitrate in 250 mL of water?
4.92 x 164 = 806.88
806.88 x 250 =201720
is this the answer???
a. To find the [H+] in the solution, we can use the pH value. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H+]). Therefore, we can determine [H+] by taking the inverse logarithm (base 10) of the negative pH value:
[H+] = 10^(-pH)
Substituting the given pH value of 2.53 into the equation:
[H+] = 10^(-2.53)
Using a calculator, we can find that [H+] is approximately 0.00352 M.
To find the percent ionization of nitrous acid, we need to make some assumptions. We assume that the concentration of [HX] at equilibrium is equal to the initial concentration. We also assume that the concentration of [H+] at equilibrium is equal to the value calculated above.
Percent ionization = ([H+] / [HX]) * 100%
Substituting the values, we can calculate the percent ionization.
b. The equilibrium expression for a weak acid, HX, can be written as follows:
HX ⇌ H+ + X-
The equilibrium constant for the dissociation of the weak acid, known as Ka, can be calculated using the concentrations of the reactants and products at equilibrium.
Ka = [H+][X-] / [HX]
We already know the concentration of [H+], which is 0.00352 M. However, we need to determine the concentration of [X-]. Since the acid is weak, we assume that the concentration of [X-] is approximately equal to the concentration of [H+]. Therefore, [X-] can also be considered as 0.00352 M.
The initial concentration of [HX] can be calculated using the given molarity:
Initial concentration (M) = Molarity * Volume (L)
Initial concentration (M) = 0.310 M * 1 L
Now, we can calculate the value of Ka:
Ka = (0.00352 M * 0.00352 M) / (0.310 M)
Using a calculator, we find that Ka is approximately 4.00 * 10^(-6).
c. To calculate the pH of the solution formed by adding 10 g of KX to 650 mL of 0.0125 M solution of the weak acid, we first need to determine the mole of KX added.
Mass = Moles * Molar mass
Moles = Mass / Molar mass
Moles = 10 g / 59.0 g•mol^(-1)
Next, we convert the volume of the weak acid solution from milliliters to liters:
Volume (L) = 650 mL * (1 L / 1000 mL)
To determine the moles of HX in the 650 mL solution, we multiply the volume (in liters) by the molarity:
Moles of HX = Volume (L) * Molarity
Moles of HX = 650 mL * (1 L / 1000 mL) * 0.0125 M
We can now calculate the new concentration of HX by adding the moles of HX from the original solution and the moles of HX added from KX:
Total moles of HX = Moles of HX (original solution) + Moles of HX (from KX)
Total moles of HX = Moles of HX + Moles of HX
Total moles of HX = 650 mL * (1 L / 1000 mL) * 0.0125 M + (10 g / 59.0 g•mol^(-1))
Finally, we can calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
Where pKa is -log(Ka) and [A-] is the concentration of X-.
d. To determine the pH of a solution formed by adding 10 g of KX to 750 mL of water, we need to first calculate the moles of KX added, as we did in part c.
Next, we convert the volume of water from milliliters to liters:
Volume (L) = 750 mL * (1 L / 1000 mL)
Now, using the moles of KX and the volume of water, we can calculate the concentration of KX in the solution:
Concentration (M) = Moles / Volume (L)
Since KX completely dissociates in water, the concentration of X- is equal to the concentration of KX.
Now, we can calculate the concentration of X-.
Using the concentration of X-, we can calculate the pOH using the equation:
pOH = -log10([X-])
Finally, we can calculate the pH using the equation:
pH = 14 - pOH