How many grams ammonium chloride and how many milliliters 3.0 M sodium hydroxide should be added to 200 ml of water and diluted to 500 ml to prepare a buffer of 9.50 with a salt concentration of 0.1 M?
Ka=9.24
I THINK this problem is asking for the amount NH4Cl to add to 3M NaOH so that YOU WILL END UP WITH 0.1M NH4^+ in 500 mL. That will be 0.05 mol NH4Cl in 500 mL.
......NH4^+ + OH^- ==> NH3 + H2O
I.....x........0........0
E...0.05.......0........y
9.50 = 9.26 log (base)/(acid)
1.82 = x/0.05
x = base = OH^- = 0.091
Therefore, if you want the final NH4^+ to base 0.05 mols (0.1M) you must start with 0.05+0.091 = 0.141 mols NH4Cl (7.54g) and add 0.091 mol NaOH.
Since M = mols/L then L = mols/M = 0.091/3M = 0.03033 or 33.33 mL 3M NaOH.
I would suggest you convert g NH4Cl to mols, convert 30.33 mL 3M NaOH to mols, set up an ICE chart and check to see if that will produce 500 mL of solution with pH = 9.50 and a final NH4Cl of 0.1M.
To prepare the buffer, you will need to calculate the amount of ammonium chloride (NH4Cl) and sodium hydroxide (NaOH) required.
Step 1: Calculate the moles of ammonium chloride (NH4Cl) needed to achieve a 0.1 M concentration.
Moles NH4Cl = Molarity (M) x Volume (L)
Moles NH4Cl = 0.1 M x 0.5 L
Moles NH4Cl = 0.05 moles
Step 2: Since the equation for the dissociation of ammonium chloride in water is NH4Cl ⇌ NH4+ + Cl-, we can see that 1 mole of NH4Cl produces 1 mole of NH4+ ions.
Therefore, the amount of NH4Cl needed in grams can be determined using its molar mass.
Molar mass of NH4Cl = 14.01 g/mol + 1.01 g/mol + 35.45 g/mol = 53.46 g/mol
Mass of NH4Cl = Moles NH4Cl x Molar mass NH4Cl
Mass of NH4Cl = 0.05 moles x 53.46 g/mol
Mass of NH4Cl = 2.673 g
So, you will need to add 2.673 grams of ammonium chloride to the 200 ml of water.
Step 3: Calculate the moles of sodium hydroxide (NaOH) needed to achieve a 0.1 M concentration.
Moles NaOH = Molarity (M) x Volume (L)
Moles NaOH = 0.1 M x 0.5 L
Moles NaOH = 0.05 moles
Step 4: The equation for the dissociation of sodium hydroxide in water is NaOH → Na+ + OH-, so 1 mole of NaOH produces 1 mole of OH- ions.
Since we want a buffer with a pH of 9.50, which is slightly basic, we will need excess OH- ions. The pOH of a solution can be calculated using the formula pOH = -log[OH-].
For pH 9.50:
pOH = 14 - pH
pOH = 14 - 9.50
pOH = 4.50
Now we need to convert the pOH to OH- concentration:
[OH-] = 10^(-pOH)
[OH-] = 10^(-4.50)
[OH-] = 3.16 x 10^(-5) M
Since 1 mole of NaOH produces 1 mole of OH- ions, we can directly use the desired OH- concentration for the calculation.
Moles NaOH needed = Moles OH- needed
Moles NaOH needed = 0.05 moles
Step 5: Calculate the volume of 3.0 M sodium hydroxide (NaOH) required.
Moles NaOH = Molarity (M) x Volume (L)
0.05 moles = 3.0 M x Volume (L)
Volume (L) = 0.05 moles / 3.0 M
Volume (L) = 0.0167 L
Since 1 L = 1000 mL, the volume of NaOH required is:
Volume (mL) = Volume (L) x 1000
Volume (mL) = 0.0167 L x 1000
Volume (mL) = 16.7 mL
So, you will need to add 2.673 grams of ammonium chloride and 16.7 mL of 3.0 M sodium hydroxide to the 200 ml of water. Then, dilute the solution to a final volume of 500 ml.
To solve this question, we will use the Henderson-Hasselbalch equation, which allows us to calculate the pH of a buffer solution:
pH = pKa + log ([A-] / [HA])
In this case, NH4Cl (ammonium chloride) is the weak acid (HA), and NaOH (sodium hydroxide) is the strong base (A-). The goal is to make a buffer solution with a pH of 9.50 and a salt concentration of 0.1 M.
Step 1: Calculate [A-] (concentration of the conjugate base)
From the Henderson-Hasselbalch equation, we have:
pH = pKa + log ([A-] / [HA])
Given that pH = 9.50 and pKa = 9.24, we can rewrite the equation as follows:
9.50 = 9.24 + log ([A-] / [HA])
Rearrange the equation to find [A-]:
log ([A-] / [HA]) = 9.50 - 9.24
log ([A-] / [HA]) = 0.26
Take the antilog of both sides to solve for ([A-] / [HA]):
[A-] / [HA] = antilog(0.26)
Step 2: Calculate the moles of NH4Cl needed
To create a solution with a final volume of 500 ml and a salt concentration of 0.1 M, we can calculate the number of moles needed using the equation:
moles = molarity × volume (in liters)
Given that the final volume is 500 ml, the final concentration is 0.1 M, and NH4Cl is the weak acid (HA), we have:
moles of NH4Cl = 0.1 M × (500 ml / 1000)
Step 3: Calculate the moles of NaOH needed
To calculate the moles of NaOH needed, we use the same equation as in Step 2:
moles = molarity × volume (in liters)
Given that the concentration is 3.0 M and the final volume is 500 ml, we have:
moles of NaOH = 3.0 M × (volume / 1000)
Step 4: Calculate the mass of NH4Cl needed
To calculate the mass of NH4Cl needed, we use the equation:
mass = moles × molar mass
The molar mass of NH4Cl can be found by adding the atomic masses of the elements: nitrogen (N), hydrogen (H), and chlorine (Cl).
Step 5: Calculate the volume of NaOH needed
Since the concentration of NaOH is given in molarity (M), we can directly use the equation:
volume = moles / molarity
Given that we have already calculated the moles of NaOH in Step 3 and the molarity is 3.0 M, we can find the volume.
Please provide the value of the final volume you want to achieve.