A 30 ml sample of of 0.150 m hydrazoic acid (ka = 4.50x10^-4 ) is titrated with a 0.100 m NaOH what is the ph after the ff additions of NaOH

30ml

45ml

60ml

1. Do you mean 0.150 m or 0.150M? You need to get your cap keys in order. I think you MUST mean M and I will work it that way. (By the way, M stands for molarity; m stands for molality.)

a. Write the equation and balance it.
b. Calculate mL NaOH to reach the equivalence point. You do that so you will know where each of these additions is on the titration curve; i.e., beginning, before or after the equivalence point or at the eq. pt. I calculate 30*0.150/0.1 = 45.0 mL NaOH but you should confirm that.

c. 30 mL is between the beginning and the eq pt; therefore you have a buffered solution. Use the Henderson-Hasselbalch equation.
d. 45 mL is at the eq pt and the pH will be determined by the hydrolysis of the salt. The cncn of the salt is 0.06 but you should confirm that. You have 30 x 0.15 = 4.5 millimols salt in 45+30 mL volume and 4.5/75 = 0.06M.
........N3^- + HOH ==> HN3 + OH^-
I.....0.06..............0.....0
C.......-x..............x.....x
E.....0.l06-x...........x.....x

Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(0.06-x) and solve for x = (OH^-). Convert that to pH.

d. At 60 mL you have passed the equivalence point. Therefore, the pH is determined by the excess NaOH.
mmols HN3 initially = 30 x 0.150 = ?
mmols NaOH = 60 x 0.1 = ?
Difference is excess OH^-.
M OH^- = mmols/mL and convert that to pOH then to pH.

I'll be happy to check your calculations if you care to post them.

To determine the pH after each addition of NaOH, we need to consider the reaction that occurs between hydrazoic acid (HN3) and sodium hydroxide (NaOH). The balanced equation for this reaction is:

HN3 + NaOH -> H2O + NaN3

We can see that for every molecule of HN3 that reacts with NaOH, one molecule of H2O is formed. The concentration of HN3 will decrease with each addition of NaOH, while the concentration of NaOH will increase.

To calculate the pH after each addition, we need to determine the concentration of HN3 remaining in solution. We can use the information given about the initial concentration (0.150 M) and the volumes (30 ml, 45 ml, 60 ml) to find the moles of HN3 initially and subtract the moles of HN3 reacted with NaOH at each step.

Step 1: 30 ml
Initial moles of HN3 = 0.150 M x 0.030 L = 0.0045 moles
Moles of HN3 reacted with NaOH = 0.100 M x 0.030 L = 0.003 moles
Moles of HN3 remaining = 0.0045 moles - 0.003 moles = 0.0015 moles

Step 2: 45 ml
Initial moles of HN3 = 0.0045 moles (from step 1)
Moles of HN3 reacted with NaOH = 0.100 M x 0.015 L = 0.0015 moles
Moles of HN3 remaining = 0.0045 moles - 0.0015 moles = 0.003 moles

Step 3: 60 ml
Initial moles of HN3 = 0.003 moles (from step 2)
Moles of HN3 reacted with NaOH = 0.100 M x 0.015 L = 0.0015 moles
Moles of HN3 remaining = 0.003 moles - 0.0015 moles = 0.0015 moles

Now that we have the moles of HN3 remaining after each addition of NaOH, we can calculate the concentration of HN3 in each step by dividing the moles by the total volume.

Step 1:
Volume = 30 ml
Molarity = moles/volume = 0.0015 moles / 0.030 L = 0.05 M
pH = -log(4.50x10^-4) = 3.35

Step 2:
Volume = 45 ml
Molarity = moles/volume = 0.003 moles / 0.045 L = 0.067 M
pH = -log(0.067) = 1.17

Step 3:
Volume = 60 ml
Molarity = moles/volume = 0.0015 moles / 0.060 L = 0.025 M
pH = -log(0.025) = 1.60

Therefore, the pH after the addition of 30 ml of NaOH is 3.35, after the addition of 45 ml is 1.17, and after the addition of 60 ml is 1.60.