What is the Ph for the solution made from 11.10mL of 0.278M HCL solution mixed with 70.90mL of 0.1300M ammonia solution? The pKa for NH4 is 9.241.
mmols HCl = 11.1 x 0.278 = 3.0858
mmoles NH3 = 70.9 x 0.13 = 9.217
..........HCl + NH3 ==> NH4Cl
initi...3.0858..9.217....0
change-3.0858..-3.0858...+3.0858
equil....0.......6.131....3.0858
From the ICE chart you can see that you have a solution of NH3 (a weak base) and its salt (NH4Cl) which makes a buffer.
Use the Henderson-Hasselbalch equation to solve for pH. Check my arithmetic.
Well, let me calculate that for you, but please don't hold your breath, as my calculations might leave you a little light-headed!
To determine the pH of the solution, we need to find the concentration of the resulting NH4+ (ammonium ion) and calculate the pOH based on that. Then we can subtract the pOH from 14 to get the pH.
First, let's find the number of moles of HCl and ammonia used:
moles of HCl = volume (L) x concentration (M)
= 11.10 mL x 0.278 M
= 0.0030878 mol
moles of ammonia = volume (L) x concentration (M)
= 70.90 mL x 0.1300 M
= 0.009217 mol
Now, let's calculate the number of moles of NH4+ formed by reacting HCl and ammonia:
1 mol of HCl reacts with 1 mol of NH3 to produce 1 mol of NH4+
So, the number of moles of NH4+ formed = 0.0030878 mol (since NH3 and HCl react in a 1:1 ratio)
Now, let's determine the concentration of NH4+:
concentration (M) = moles / volume (L)
= 0.0030878 mol / (11.10 mL + 70.90 mL) [converted to liters]
= 0.0030878 mol / 0.082 L
= 0.0377 M
Now that we have the concentration of NH4+, let's calculate the pOH using the pKa value for NH4:
pOH = pKa + log(concentration of NH4+ / concentration of NH3)
= 9.241 + log(0.0377 M / 0.1300 M) [concentration of NH3 is given, so we don't need to calculate it]
= 9.241 + log(0.2892)
= 9.241 + (-0.537)
= 8.704
Finally, we can find the pH by subtracting the pOH from 14:
pH = 14 - pOH
= 14 - 8.704
= 5.296
So, the pH of the solution made from mixing 11.10 mL of 0.278 M HCl solution with 70.90 mL of 0.1300 M ammonia solution is approximately 5.296.
To find the pH of the solution, we will first calculate the concentration of the resulting NH4Cl solution by using the principles of stoichiometry and the balanced chemical equation:
HCl + NH3 -> NH4Cl
Step 1: Calculate the moles of HCl:
moles of HCl = volume of HCl solution (in L) x concentration of HCl solution (in mol/L)
moles of HCl = 11.10 mL x (1 L/1000 mL) x 0.278 mol/L
moles of HCl = 0.0030838 mol
Step 2: Calculate the moles of NH3:
moles of NH3 = volume of NH3 solution (in L) x concentration of NH3 solution (in mol/L)
moles of NH3 = 70.90 mL x (1 L/1000 mL) x 0.1300 mol/L
moles of NH3 = 0.009217 mol
Step 3: Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed. In this case, the limiting reactant will be NH3 because it has fewer moles than HCl. This means that all of the NH3 will react with HCl, leaving some excess HCl.
Step 4: Calculate the moles of NH4Cl formed:
The balanced chemical equation shows that one mole of NH3 reacts with one mole of HCl to form one mole of NH4Cl. Since NH3 is the limiting reactant, the moles of NH4Cl formed will be equal to the moles of NH3.
moles of NH4Cl = 0.009217 mol
Step 5: Calculate the concentration of NH4Cl:
volume of NH4Cl solution = volume of HCl solution + volume of NH3 solution
volume of NH4Cl solution = 11.10 mL + 70.90 mL
volume of NH4Cl solution = 82 mL
concentration of NH4Cl = moles of NH4Cl / volume of NH4Cl solution
concentration of NH4Cl = 0.009217 mol / (82 mL x 1 L/1000 mL)
concentration of NH4Cl = 0.1124 mol/L
Step 6: Calculate the concentration of NH4+ (ammonium ion):
Since NH4Cl is fully dissociated in water, the concentration of NH4+ will be equal to the concentration of NH4Cl.
concentration of NH4+ = 0.1124 mol/L
Step 7: Calculate the concentration of OH- (hydroxide ion):
Since NH3 is a weak base, it reacts with water to produce OH- ions. The concentration of OH- can be calculated using the Kb expression for NH3:
Kb = [NH4+][OH-] / [NH3]
Kb = 1.0 x 10^-14 / 9.241
Now, rearranging the equation to calculate [OH-]:
[OH-] = (Kb x [NH3]) / [NH4+]
[OH-] = (1.0 x 10^-14 / 9.241) x (0.009217 mol / 0.1124 mol/L)
[OH-] = 1.091 x 10^-3 mol/L
Step 8: Calculate the pOH of the solution:
pOH = -log[OH-]
pOH = -log(1.091 x 10^-3)
pOH = 2.962
Step 9: Calculate the pH of the solution:
pH = 14 - pOH
pH = 14 - 2.962
pH = 11.038
Therefore, the pH of the solution made from 11.10 mL of 0.278 M HCl solution mixed with 70.90 mL of 0.1300 M ammonia solution is approximately 11.038.
To find the pH of the solution, we need to consider the reaction between HCl and ammonia. HCl is a strong acid, while ammonia (NH3) is a weak base that can react with water to form ammonium ions (NH4+) and hydroxide ions (OH-).
The balanced chemical equation for the reaction is:
HCl + NH3 → NH4+ + Cl-
To determine the pH, we need to calculate the concentrations of the reactants and products involved in the reaction.
Step 1: Calculate the moles of HCl and NH3
Moles of HCl = volume (in L) × concentration
Moles of HCl = 0.01110 L × 0.278 mol/L = 0.0030838 mol
Moles of NH3 = volume (in L) × concentration
Moles of NH3 = 0.07090 L × 0.1300 mol/L = 0.009217 mol
Step 2: Determine the limiting reactant
In the balanced equation, the mole ratio between HCl and NH3 is 1:1. Since the moles of HCl (0.0030838 mol) are less than the moles of NH3 (0.009217 mol), HCl is the limiting reactant.
Step 3: Calculate the moles of NH4+ formed
Since the mole ratio between HCl and NH4+ is 1:1, the moles of NH4+ will be equal to the moles of HCl.
Moles of NH4+ = 0.0030838 mol
Step 4: Calculate the concentration of NH4+
Concentration of NH4+ = moles/volume
Concentration of NH4+ = 0.0030838 mol / (0.01110 L + 0.07090 L) = 0.0285 mol/L
Step 5: Calculate the pOH
pOH = -log10 [OH-]
From the balanced equation, we know that 1 mole of NH4+ and 1 mole of OH- are formed for every mole of HCl. Therefore, the concentration of OH- is the same as the concentration of NH4+.
Concentration of OH- = 0.0285 mol/L
pOH = -log10 (0.0285) = 1.545
Step 6: Calculate the pH
pH + pOH = 14
pH + 1.545 = 14
pH = 14 - 1.545
pH = 12.455
Therefore, the pH of the solution made from the given volumes and concentrations is approximately 12.455.