calculate the amount of solid NH4Cl to add to 50 mL of 1 M to form a mixture that has a pH of 9 assume no volume change in the solution
----to add to 1 M WHAT to form a mixture====
oops ammonia
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
To calculate the amount of solid NH4Cl to add to 50 mL of 1 M solution to achieve a pH of 9, we first need to understand the chemical reactions that occur when NH4Cl dissolves in water.
NH4Cl dissociates in water to form NH4+ and Cl- ions:
NH4Cl(s) → NH4+(aq) + Cl-(aq)
In water, the NH4+ ion can act as an acid, while the Cl- ion remains as a spectator ion. The NH4+ ion can donate a proton (H+) to water, resulting in the formation of NH3 (ammonia) and H3O+ (hydronium ion):
NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq)
The equilibrium expression for this reaction is represented by the acid dissociation constant, Ka, which is given by:
Ka = [NH3][H3O+]/[NH4+]
Since we want the mixture to have a pH of 9, which corresponds to a hydronium ion concentration of 1 x 10^(-9) M, we can assume that the concentration of NH3 (ammonia) formed will also be 1 x 10^(-9) M.
Now, let's determine how much NH4Cl we need to add to achieve this concentration. We know that the initial volume is 50 mL (0.050 L) and the concentration is 1 M (1 mol/L). Therefore, the number of moles of NH4Cl in the solution can be calculated using the equation:
moles of NH4Cl = concentration × volume
= 1 mol/L × 0.050 L
= 0.050 moles NH4Cl
Since the NH4+ ion reacts with water to form NH3 and H3O+, we need to take into account the stoichiometry of this reaction. From the balanced equation, we can see that for every one NH4+ ion, one NH3 molecule is formed. Therefore, the moles of NH4+ ion are also equal to 0.050.
The concentration of NH4+ ion in the final solution should be equal to the concentration of NH3, which is 1 x 10^(-9) M.
Using the volume of the final solution, which is 50 mL (0.050 L), we can calculate the moles of NH4+ ion required:
moles of NH4+ = concentration × volume
= 1 x 10^(-9) mol/L × 0.050 L
= 5 x 10^(-11) moles NH4+
Since both NH4Cl and NH4+ have a 1:1 stoichiometric ratio, we need 5 x 10^(-11) moles of NH4Cl to achieve a pH of 9 in the 50 mL solution.
In summary, you would need to add approximately 5 x 10^(-11) moles of NH4Cl to 50 mL of 1 M solution to form a mixture with a pH of 9, assuming no volume change.