Calculate the pH of each of the solutions and the change in pH to 0.01 pH units

Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.90-M NaOH to 610. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative. a) water pH before mixing = pH after mixing= pH change = b) 0.163 M NH4^1+ pH before mixing = pH after mixing= pH change = c) 0.164 M NH3 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.163 M in each NH4^1+ and NH3 pH before mixing = pH after mixing= pH change =

Explanation / Answer

Solution :-

Volume = 610 ml = 0.610 L

Volume of NaOH = 10 ml and molarity is 3.90 M

a)Water

pH before mixing = 7

pH after mixing = 12.8

pH change = (12.8-7)= 5.80

to calculate the pH after mixing we need to calculate the new molarity of the NaOH

New molarity of the NaOH = 10 ml * 3.90 M /620 ml

                                                = 0.0629 M

Now lets calculate the pOH

pOH= -log[OH-]

pOH= -log [0.0629]

pOH= 1.20

now

pH+ pOH= 14

pH= 14- pOH

     = 14-1.20

    = 12.80

b) 0.163 M NH4+

initial pH can be calculated using the ka of the NH4+

ka = kw / kb

ka= 1*10^-14 / 1.8*10^-5

ka= 5.56*10^-10

so

ka= [H3O+] [NH3]/[NH4+]

5.56*10^-10 = [x][x]/[0.163]

5.56*10^-10 * 0.163 = x^2

9.06*10^-11 = x^2

Taking square root of both sides we get

9.52*10^-6 = [x]=[H+]

pH= -log [H+]

pH= - log [9.52*10^-6]

pH= 5.02

Calculating the pH after mixing

moles of NH4+ = 0.163 mol per L * 0.610 L = 0.09943 mol

moles of NaOH = 3.90 mol per L * 0.010 L = 0.039 mol

so after the reaction moles of NH4+ remain = 0.09943 mol – 0.039 mol =0.06043 mol

moles of NH3 formed = 0.039 mol

now using the Henderson equation lets calculate the pH of the solution after mixing

pOH = pkb + log [Acid /base]

pOH= 4.74 + log [0.06043 /0.039]

pOH= 4.93

pH= 14 – pH

pH= 14 – 4.93

pH= 9.07

Change in pH = 9.07 – 5.02 = 4.05

C) 0.163 M NH3

Moles of NH3 = 0.163 mol per L * 0.610 L = 0.09943 mol

Moles of NaOH = 0.039 mol

pH before mixing

kb of the NH3 = 1.8*10^-5

kb = [NH4+] [OH-] / [NH3]

1.8*10^-5 = [x][x]/[0.163]

1.8*10^-5 *0.163 =x^2

2.934 *10^-6 = x^2

Taking square root of both sides we get

X=0.00171 = [OH-]

pOH= -log [OH-]

pOH= -log [0.00171]

pOH= 2.77

pH= 14 – 2.77 = 11.23

After adding the NaOH

Concentration of the OH- = 10 ml *3.9 M / 620 ml = 0.0629 M

So the pOH = -log [0.0629]

pOH = 1.20

pH= 14 – pOH

pH= 14-1.20 = 12.80

Change in pH = 12.80-11.23 = 1.57

d)

pH before mixing

pH= 9.26 + log [base / acid ]

pH= 9.26 + log [0.163/0.163]

pH= 9.26

pH after mixing

after mixing moles of NH4+ = 0.09943 mol -0.039 mol = 0.06043 mol

moles of NH3 = 0.09943 mol + 0.039 mol = 0.13843 mol

pH= pka + log [base/acid]

pH= 9.26 + log [0.13843 /0.06043]

pH = 9.46

Change in the pH= 9.46 – 9.26 = 0.200

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