Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH.

For 270.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH.

For 270.0 mL of a buffer solution that is 0.240 *M* in
HCHO2 and 0.310 *M* in KCHO2, calculate the initial pH and
the final pH after adding 0.0150 mol of NaOH(Ka=1.8⋅10−4).

For 270.0 mL of a buffer solution that is 0.2604 *M* in
CH3CH2NH2 and 0.2304 *M* in CH3CH2NH3Cl, calculate the
initial pH and the final pH after adding 0.0150 mol of
NaOH(Kb=5.6⋅10−4).

Answer #1

1)

i) Initial **pH of pure water** will be
**7.**

ii)NaOH on addition to water dissociates as, NaOH(s) + H2O
------------- > Na+(aq.) + OH^{-}(aq.)

For aq.NaOH solution formed,

V=270.0 mL, [NaOH] = C = 0.0150 M, [OH^{-}] =?

NaOH is strong electrolyte which dissociates to the complete extent and hence,

[OH^{-}] = [NaOH] = 0.0150 M

Let us calculate the concentration in terms of M/L

0.0150 M of NaOH added to 270 mL of water, hence,

[OH^{-}] = 0.0150 M / 270 mL

So, [OH^{-}] = 5.56 x 10^{-5}
M/**mL**

So, [OH^{-}] = (5.56 x 10^{-5}) x 10^{3}
M/**L**

[OH^{-}] = 5.56 x 10^{-2} M/L ……………………… (Anathor
way -From this we can find pOH and subtract it from 14 to get
pH)

We know that,

[OH^{-}][H^{+}] = Kw = 1 x 10^{-14}

[OH^{-}][H^{+}] = 1 x 10^{-14}

(5.56 x 10^{-2}) [H^{+}] = 1 x
10^{-14}

[H^{+}] = (1 x 10^{-14}) / (5.56 x
10^{-2})

[H^{+}] = 1.8 x 10-13

Now, pH = -log [H^{+}]

pH = -log(1.8x 10^{-13})

**pH = 12.75**

=============================================

2)

i)Let initial pH be pH1.

For this solution [CHO2^{‑}] : [HCHO2] = 0.31: 0.24,

Ka = 1.8 x 10-4

So, pKa = -log(ka)

pKa = - log(1.8 x 10^{-4})

pKa = 3.7447

According to Henderson’s equation,

pH = pKa + log{[conjugate base or salt]/[acid]}

Hence here we have,

pH1 = pKa + log{[CHO2^{-}]/[HCHO2]}

pH1 = 3.7447 + log(0.31/0.24)

pH1 = 3.7447 + 0.1112

pH1 = 3.8559

ii)Let pH after adding NaOH be pH2.

Addition of strong base like NaOH decrease the concentration of acid and increases the concentration of conjugate base (i.e. salt)

Let us find new concentrations,

Concentration of NaOH = 0.0150 x 270 = 4.05 mmol of NaOH

Originally, [HCHO2] = 0.240x270 = 64.8 mmol

After adding 4.05 mmol NaOH, [HCHO2] = 64.8 – 4.05 = 60.75

Originally, [KCHO2] = 0.310 x 270 = 83.7 mmol

After adding NaOH [CHO_{2}^{-}] = 83.7 + 4.05 =
87.75 mmol.

Hence new, [CHO2^{‑}] : [HCHO2] = 87.75 :
60.75

By Henderson’s equation,

pH2 = 3.7447 + log(87.75/60.75)

pH2 = 3.7447 + 0.1597

pH2 = 3.9044

==========================================================

3)

Given buffer is Ethyl amine (EA)/ethylammoniumchloride(EAcl) and it’s a basic buffer. We will calculate first pOH and then pH.

i) Let initial pH be pH1,

We have, Kb = 5.6 x 10-4

so, pKb = -log (Kb) = 3.2518

[EAcl]:[EA] = 0.2304:0.2604

Henderson’s equation for Basic buffer,

pOH = pKb + log{[conjugate acid]/[Base]}

pOH = pKb + log{[EAcl]/[EA]}

pOH1 = 3.2518 + log(0.2304/0.2604)

pOH1 = 3.2518 + (-0.0532)

pOH1 = 3.1986

hence,

pH1 = 14 – pOH1

pH1 = 14 – 3.1986

**pH1 = 10.8014.**

ii) After adding strong base NaOH, [EAcl] will decrease and that of [EA] will increase.

Let us find new concentrations, after adding 0.0150 M to 270 mL (i.e. 4.05 mmol) of buffer.

Originally, [EAcl] = 0.2304 x 270 = 62.21 mmol

After adding NaOH, [EAcl] = 62.21 – 4.05 = 58.16 mmol.

Originally [EA] = 0.2604 x 270 = 70.31 mmol.

After adding NaOH [EA] = 70.31 + 4.05 = 74.36 mmol

Hence, new [EAcl] : [EA] = 58.16:76.36

Henderson’s equation,

pOH2 = pKb + log(58.16/76.36)

pOH2 = 3.2518 + (-0.1182)

pOH2 = 3.1336

pH2 = 14 – 3.1336

pH2 = 14-3.6265

**pH2 = 10.8664**

**======================XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX=========================**

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.0100 mol of NaOH.
Part A
For 300.0 mL of pure water, calculate the initial pH and the
final pH after adding 0.0100 mol of NaOH.
Express your answers using two decimal places separated by a
comma.
Part B
For 300.0 mL of a buffer solution that is 0.210 M in
HCHO2 and 0.280 M in KCHO2, calculate the initial pH and
the final...

For each of the following solutions, calculate the
initial pH and the final pH after adding 0.0200 mol of
NaOH.
Part A
For 290.0 mL of pure water, calculate the initial pH and the
final pH after adding 0.0200 mol of NaOH.
Express your answers using two decimal places separated by a
comma.
Answer: pHinitial, pHfinal = 7.00,12.84
Part B
For 290.0 mL of a buffer solution that is 0.200 M in
HCHO2 and 0.310 M in KCHO2, calculate the...

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.010 mol of NaOH.
Part A: For 270.0 mL of a buffer solution that is 0.205 M in
HCHO2 and 0.305 M in KCHO2, calculate the initial pH and the final
pH after adding 0.010 mol of NaOH. Express your answers using two
decimal places separated by a comma.
Part B: For 270.0 mL of a buffer solution that is 0.285 M in
CH3CH2NH2...

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.010 mol of NaOH.
For 300.0mL of pure water, calculate the initial pH and the
final pH after adding 0.010 mol of NaOH.
For 300.0mL of a buffer solution that is 0.240M in
HCHO2 and 0.280M in KCHO2, calculate the initial pH and
the final pHafter adding 0.010 mol of NaOH.
For 300.0mL of a buffer solution that is 0.305M in
CH3CH2NH2 and 0.285M...

Exercise 16.50
For each of the following solutions, calculate the initial pH
and the final pH after adding 0.0200 mol of NaOH.
Part A
For 300.0 mL of pure water, calculate the initial pH and the
final pH after adding 0.0200 mol of NaOH.
Express your answers using two decimal places separated by a
comma.
pHinitial, pHfinal =
7.00,12.82
SubmitMy AnswersGive
Up
Correct
Part B
For 300.0 mL of a buffer solution that is 0.210 M in
HCHO2 and 0.310...

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.010 mol of NaOH .
Part B
250.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.305
M in KCHO2
Part C
250.0 mL of a buffer solution that is 0.265 M in CH3CH2NH2 and
0.245 M in CH3CH2NH3Cl
Express your answers using two decimal places separated by a
comma.

For each of the following solutions, calculate the initial pHand
the final pH after adding 0.005 mol of NaOH.
260.0 mL of a buffer solution that is 0.225 M in HCHO2 and 0.315
M in KCHO
260.0 mL of pure water
260.0 mL of a buffer solution that is 0.275 M in CH3CH2NH2 and
0.245 M in CH3CH2NH3Cl

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.005 mol of NaOH.
For 260.0 mL of a buffer solution that is 0.265 M in
CH3CH2NH2 and 0.235 M in CH3CH2NH3Cl, calculate the
initial pH and the final pH after adding 0.005 mol of NaOH.

for each of the following Solutions calculate the
initial pH in the final pH after adding 0.200 moles of NaOH for
300.0 M L of a buffer solution that is 0.2551 m in
ch3ch2nh2 AND 0.2351 in ch3ch2nh3cl calculate the initial pH and
the final pH after adding 0.020 moles of naoh KB equal 5.6 x 10 to
the negative 4

For each of the following solutions, calculate the initial pH
and the final pH after adding 0.010 mol of NaOH.
c) For 220.0 mL of a buffer solution that is 0.285 M in
CH3CH2NH2 and 0.255 M in CH3CH2NH3Cl, calculate the
initial pH and the final pH after adding 0.010 mol of NaOH.

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