- #1

- 24

- 2

- Homework Statement
- Calculate [OH- ] in a solution obtained by adding 0.0100 mol

solid NaOH to 1.00 L of 15.0 M NH3 .

- Relevant Equations
- Is the solution manual goes wrong? Or I am confused by my self? please educate me

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Chemistry
- Thread starter r12214001
- Start date

- #1

- 24

- 2

- Homework Statement
- Calculate [OH- ] in a solution obtained by adding 0.0100 mol

solid NaOH to 1.00 L of 15.0 M NH3 .

- Relevant Equations
- Is the solution manual goes wrong? Or I am confused by my self? please educate me

- #2

Mentor

- 29,257

- 3,952

- #3

Homework Helper

Gold Member

- 3,963

- 1,002

It's a good question. I could not read the student's solution but my first calculation gave me exactly the same result as his so we must have been thinking the same.

Generic thoughts that are usually valid but not quite in this case are: if you get quadratics in a problem of this kind it is because you have overlooked some simplifying approximation; or that buffering power is maximum at pH = pK_{a} (= 9.255) while here we are far away and on the very edge where buffering power is in principle zero (the pH of 15 M ammonia solution is 12.21). But actually normal thoughts about buffering are a bit out of place here because normally we would be thinking about what happens when we add **strong acid** to an ammonia solution not strong base.

The precise error is we both assumed that the molarity (0.01 M) of added NaOH equal the increase of [OH^{-}] which is not quite right in the precise conditions (high ammonia concentration, low amount of added Na^{+}). But what is happening is that while a high concentration of ammonia solution is mostly NH_{3} still there is a small amount of NH_{4}^{+} - 0.00424 M and 0.00164 M for 1M and 15 M total ammonia respectively*. So as strong base is added these small amounts are deprotonated particularly within the first small amounts added and the Δ[OH^{-}] < Δ[Na^{+}]. That is like buffering, but this is not what we normally think of as buffering, or if it is it is by buffer of very low concentration.

**(These concentrations are calculated from the approximate equation ##\left[ OH^{-}\right] =\sqrt {N_{T}K_{b}}##. A more exact calculation is solving the quadratic below for [Na*^{+}] = 0, but the two agree to the third decimal place.)

Thus solving the quadratic (which we don't really need to do, see below)

$$x^{2}+x\left( K_{b}-\left[ Na^{+}\right] \right) -K_{b}\left( N_{T}+\left[ Na^{+}\right] \right) =0$$

where N_{T} is total concentration of the weak base and ##x## is [OH^{-}], we find that on adding 0.01 M NaOH Δ[OH^{-}] = 0.0072 < 0.01. This pseudobuffering soon runs out: on a second such addition Δ[OH^{-}] = 0.0093 and on a third 0.00973, pretty close to the NaOH added. At that point pH = 12.49 by the way, everything here is happening in a small pH range. At !5 M ammonia I calculate for the first ∆[Na^{+}] of 0.01, ∆[OH^{-}] is as low as 0.0057, an appreciable pseudobuffering effect for the somewhat special and narrow-range conditions chosen.

I take it as understood how the quadratic equation above is derived from the equilibrium equation, a mass conservation equation and an 'electroneutrality' condition with the excellent approximation ##\left[ H^{+}\right] =0##. For understanding and even most quantitative solutions this quadratic is not that necessary nor helpful and better for understanding is its inverse

$$\left[ Na^{+}\right] =x+\dfrac {K_{b}N_{T}}{x+K_{b}}$$

Here the corresponding curves for N_{T} = 1 (blue) and 15 M (purple).

(Vertical co-ordinate [Na^{+}], horizontal distances between any two points represent Δ[OH^{-}] - note that pure ammonia solution [OH^{-}] is that where the curve for the corresponding N_{T} intersects the abscissa.)

We see that after an early deviation they soon approximate to the line [Na^{+}] = ##x##, more slowly with the massive 15 M concentration.

Talking of which all this this is all very well for fine understanding of our equations, but I would think at 15 M ammonia the dielectric constant of the solution would have changed the given K_{b} quite a lot from the one given, that will have been determined in a more dilute solutoon.

Generic thoughts that are usually valid but not quite in this case are: if you get quadratics in a problem of this kind it is because you have overlooked some simplifying approximation; or that buffering power is maximum at pH = pK

The precise error is we both assumed that the molarity (0.01 M) of added NaOH equal the increase of [OH

*

Thus solving the quadratic (which we don't really need to do, see below)

$$x^{2}+x\left( K_{b}-\left[ Na^{+}\right] \right) -K_{b}\left( N_{T}+\left[ Na^{+}\right] \right) =0$$

where N

I take it as understood how the quadratic equation above is derived from the equilibrium equation, a mass conservation equation and an 'electroneutrality' condition with the excellent approximation ##\left[ H^{+}\right] =0##. For understanding and even most quantitative solutions this quadratic is not that necessary nor helpful and better for understanding is its inverse

$$\left[ Na^{+}\right] =x+\dfrac {K_{b}N_{T}}{x+K_{b}}$$

Here the corresponding curves for N

(Vertical co-ordinate [Na

We see that after an early deviation they soon approximate to the line [Na

Talking of which all this this is all very well for fine understanding of our equations, but I would think at 15 M ammonia the dielectric constant of the solution would have changed the given K

Last edited:

- #4

- 24

- 2

Being surprised with tour detailed inteprenetion TKSIt's a good question. I could not read the student's solution but my first calculation gave me exactly the same result as his so we must have been thinking the same.

Generic thoughts that are usually valid but not quite in this case are: if you get quadratics in a problem of this kind it is because you have overlooked some simplifying approximation; or that buffering power is maximum at pH = pK_{a}(= 9.255) while here we are far away and on the very edge where buffering power is in principle zero (the pH of 15 M ammonia solution is 12.21). But actually normal thoughts about buffering are a bit out of place here because normally we would be thinking about what happens when we addstrong acidto an ammonia solution not strong base.

The precise error is we both assumed that the molarity (0.01 M) of added NaOH equal the increase of [OH^{-}] which is not quite right in the precise conditions (high ammonia concentration, low amount of added Na^{+}). But what is happening is that while a high concentration of ammonia solution is mostly NH_{3}still there is a small amount of NH_{4}^{+}- 0.00424 M and 0.00164 M for 1M and 15 M total ammonia respectively*. So as strong base is added these small amounts are deprotonated particularly within the first small amounts added and the Δ[OH^{-}] < Δ[Na^{+}]. That is like buffering, but this is not what we normally think of as buffering, or if it is it is by buffer of very low concentration.

*(These concentrations are calculated from the approximate equation ##\left[ OH^{-}\right] =\sqrt {N_{T}K_{b}}##. A more exact calculation is solving the quadratic below for [Na^{+}] = 0, but the two agree to the third decimal place.)

Thus solving the quadratic (which we don't really need to do, see below)

$$x^{2}+x\left( K_{b}-\left[ Na^{+}\right] \right) -K_{b}\left( N_{T}+\left[ Na^{+}\right] \right) =0$$

where N_{T}is total concentration of the weak base and ##x## is [OH^{-}], we find that on adding 0.01 M NaOH Δ[OH^{-}] = 0.0072 < 0.01. This pseudobuffering soon runs out: on a second such addition Δ[OH^{-}] = 0.0093 and on a third 0.00973, pretty close to the NaOH added. At that point pH = 12.49 by the way, everything here is happening in a small pH range. At !5 M ammonia I calculate for the first ∆[Na^{+}] of 0.01, ∆[OH^{-}] is as low as 0.0057, an appreciable pseudobuffering effect for the somewhat special and narrow-range conditions chosen.

I take it as understood how the quadratic equation above is derived from the equilibrium equation, a mass conservation equation and an 'electroneutrality' condition with the excellent approximation ##\left[ H^{+}\right] =0##. For understanding and even most quantitative solutions this quadratic is not that necessary nor helpful and better for understanding is its inverse

$$\left[ Na^{+}\right] =x+\dfrac {K_{b}N_{T}}{x+K_{b}}$$

Here the corresponding curves for N_{T}= 1 (blue) and 15 M (purple).

(Vertical co-ordinate [Na^{+}], horizontal distances between any two points represent Δ[OH^{-}] - note that pure ammonia solution [OH^{-}] is that where the curve for the corresponding N_{T}intersects the abscissa.)

View attachment 255471

View attachment 255476

We see that after an early deviation they soon approximate to the line [Na^{+}] = ##x##, more slowly with the massive 15 M concentration.

Talking of which all this this is all very well for fine understanding of our equations, but I would think at 15 M ammonia the dielectric constant of the solution would have changed the given K_{b}quite a lot from the one given, that will have been determined in a more dilute solutoon.

Share:

- Replies
- 2

- Views
- 372

- Replies
- 2

- Views
- 1K

- Replies
- 9

- Views
- 748

- Replies
- 4

- Views
- 372

- Replies
- 1

- Views
- 270

- Replies
- 2

- Views
- 437

- Replies
- 3

- Views
- 2K

Chemistry
I forgot how to do chemistry math

- Replies
- 6

- Views
- 994

- Replies
- 2

- Views
- 2K

- Replies
- 2

- Views
- 3K