what is the pH for the solution made from 11.1 mL of 0.278M NaOH solution mixed with 70.2mL of 0.13M ammonium chloride solution? pKa for NH4+ is 9.241
To find the pH of the solution, we need to consider the reaction between the NaOH and NH4Cl. NaOH is a strong base and NH4Cl is a salt of a weak acid (NH4+) and its conjugate base (Cl-). This reaction will result in the formation of NH3 (ammonia) and H2O:
NH4Cl + NaOH → NH3 + H2O + NaCl
First, let's calculate the moles of NaOH and NH4Cl used in the solution:
Moles of NaOH = volume (L) × concentration (mol/L)
Moles of NaOH = 0.0111 L × 0.278 mol/L
Moles of NaOH = 0.0030818 mol
Moles of NH4Cl = volume (L) × concentration (mol/L)
Moles of NH4Cl = 0.0702 L × 0.13 mol/L
Moles of NH4Cl = 0.009126 mol
Next, let's determine which reactant is the limiting reactant. The limiting reactant is the one that runs out first and determines the extent of the reaction.
In this case, NaOH is the limiting reactant because it has fewer moles than NH4Cl.
Since all the NaOH will react, we can calculate the moles of NH3 formed using stoichiometry. From the balanced equation, we know that the ratio of NaOH to NH3 is 1:1:
Moles of NH3 formed = Moles of NaOH used = 0.0030818 mol
Now, let's calculate the moles of NH4+ remaining after the reaction:
Moles of NH4+ remaining = Moles of NH4Cl used - Moles of NH3 formed
Moles of NH4+ remaining = 0.009126 mol - 0.0030818 mol
Moles of NH4+ remaining = 0.0060442 mol
Now, we can determine the concentration of NH4+ in the final solution:
Concentration of NH4+ = Moles of NH4+ remaining / volume of final solution (L)
Concentration of NH4+ = 0.0060442 mol / (0.0111 L + 0.0702 L)
Concentration of NH4+ = 0.0060442 mol / 0.0813 L
Concentration of NH4+ = 0.074380 mol/L
To calculate the pOH, we need to find the concentration of OH- in the final solution:
Concentration of OH- = Concentration of NaOH (0.278 mol/L) / Total volume (L)
Concentration of OH- = 0.0030818 mol / (0.0111 L + 0.0702 L)
Concentration of OH- = 0.0030818 mol / 0.0813 L
Concentration of OH- = 0.037920 mol/L
pOH = -log[OH-]
pOH = -log(0.037920)
pOH = 1.42
Finally, let's calculate the pH using the pOH and the pKa value given:
pH = 14 - pOH - pKa
pH = 14 - 1.42 - 9.241
pH = 3.339
Therefore, the pH of the solution is approximately 3.339.
To determine the pH of the solution, you need to consider the reaction between sodium hydroxide (NaOH) and ammonium chloride (NH4Cl). NaOH is a strong base, and NH4Cl is a salt of a weak acid (NH4+) and a strong base (Cl-).
First, calculate the amount of moles of NaOH and NH4Cl in the respective solutions.
For NaOH:
Moles of NaOH = volume (in liters) × concentration in M
Moles of NaOH = 11.1 mL × (1 L / 1000 mL) × 0.278 M
For NH4Cl:
Moles of NH4Cl = volume (in liters) × concentration in M
Moles of NH4Cl = 70.2 mL × (1 L / 1000 mL) × 0.13 M
Next, determine which of the two reactants will be completely consumed in the reaction. Since NaOH is in excess, it will react completely with NH4Cl according to the equation:
NH4Cl + NaOH → NH3 + H2O + NaCl
The moles of NH4Cl reacted with NaOH are equal to the initial moles of NH4Cl:
Moles of NH4Cl reacted = Moles of NH4Cl
Now, calculate the amount of moles of NH3 produced:
Moles of NH3 = Moles of NH4Cl reacted
Next, calculate the concentration of NH3 (NH4+) in the solution using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
In this case, [A-] represents the concentration of NH3 (NH4+) formed, and [HA] is the concentration of NH4Cl initially.
[A-] = Moles of NH3 / Total volume of solution
[HA] = Moles of NH4Cl / Total volume of solution
Finally, substitute the values into the Henderson-Hasselbalch equation to find the pH.
Please follow the steps and calculations mentioned above to find the pH of the given solution.