Back titration is also titration



• What is pH of 0.0002 M HI solution? What will be pH if 1 mL of the above was added to 999 mL of pure water?

• What is pH of mixture prepared by mixing 20 mL 0.07 M NaOH and 13 mL 0.09 M HCl?

• Calculate pH of 1 M solution of nitrous acid, pKa=3.37.

• Calculate pH of 0.001 M solution of benzoic acid, pKa=4.19.

• 0.1 M solution of weak acid has pH=4.0. Calculate pKa.

• 20 mL of 0.1 M solution of weak acid was mixed with 8 mL 0.1 M solution of NaOH. Measured pH was 5.12. Calculate pKa.

• Calculate the pH of a 0.2M solution of ammonia. Ka=5.62×10-10.

• Calculate pH of 0.01 M aniline hydrochloride. Aniline pKb=9.4.

• Calculate pH of 0.1 M potassium hydrogen oxalate. pKa1=1.25, pKa2=4.27.

• Calculate pH of solution prepared from 0.1 mole of formic acid and 0.02 mole of NaOH diluted to 1L. PKa=3.75.

• You have 100 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare

What is pH of 0.0002M HI solution?

What will be pH if 1 mL of the above was added to 999 mL of pure water?

HI is a strong acid so we can assume it is fully dissociated.

3×10-4 M solution is concentrated enough to allow us to neglect water dissociation (see table in strong acid section). Thus we can use 7.6 equation

[pic]7.6

and the pH calculated is -log(3×10-4)=3.70.

After the solution has been diluted 1000 times HI concentration is 2×10-7 M and we are in the area where water dissociation may spoil results of the 7.6 equation. Thus it wil be better to use the equation 7.4

[pic]7.4

and calculated pH=6.62 (you know how to solve quadratic equation, don't you?).

Using equation 7.6 we will get pH=6.70. Difference is not large - but clearly visible.

As additional confirmation our pH calculator shows exactly the same results - 3.70 and 6.62.

What is pH of mixture prepared by mixing 20 mL 0.07 M NaOH and 13 mL 0.09 M HCl?

Although you are asked about pH, this is a simple limiting reagent question.

When you mix strong acid and strong base they react till one of the reagents runs out. What is left is responsible for pH.

Reaction taking place in the solution is

NaOH + HCl -> NaCl + H2O

Initially there were 20×0.07=1.4 mmol of NaOH and 13×0.09=1.17 mmol of HCl. They react 1:1, so after reaction 1.4-1.17=0.23 mmol of NaOH is left. Final volume of the solution is 20+13=33 mL, thus final concentration of NaOH left is 0.23/33=0.0070 M. pOH=-log(0.0070)=2.15 and pH=14-2.15=11.85.

Note that if there is exactly the same amount of both reactants reaction proceeds to the end - and you are left with perfectly neutral solution of pH=7.00.

Calculate pH of 1M solution of nitrous acid, pKa=3.37.

Let's try first with the simplified formula 8.13.

[pic]8.13

If we put concentration and Ka values into formula we get [H+]=0.021 M and pH=1.69. Is it correct? To be sure we have to check if the 5% rule is obeyed, that is if the dissociation fraction is less then 5%.

[pic]1.3

Dissociated acid concentration is the same as [H+] - if we assume we can neglect water dissociation. It is obvious that we can, as 0.021 M is five orders of magnitude larger than the concentration of H+ from the water. So dissociation fraction is 0.021/1 - a little bit over 2%, much less then 5%, so the 5% rule is obeyed.

As additional confirmation our pH calculat

Calculate pH of 0.001M solution of benzoic acid, pKa=4.19.

We will start with the simplified formula 8.13.

[pic]8.13

From formula we get [H+]=2.54×10-4 M and pH=3.60. Is it correct? To be sure we have to check if the 5% rule is obeyed, that is if the dissociation fraction is less then 5%.

[pic]1.3

Dissociated acid concentration is the same as [H+], thus dissociation fraction is 2.54×10-4/0.001 - above 25%. We can't use simplified formula 8.13! We have to use more precise equation 8.10:

[pic]8.10

This time we get pH=3.65, exactly the

0.1 M solution of weak acid has pH=4.0. Calculate pKa.

If pH=4.0 then [H+]=10-4 - that means that dissociation fraction is about

10-4/0.1×100%=0.1% - well below 5%. That in turn means that pH of the solution is described by the simplified formula 8.13.

[pic]8.13 All we have to do is rearrange to get equation for Ka:

[pic]

and Ka=(10-4)2/0.1=10-7 and pKa=7.

Our pH calculator can help us check this result - and indeed, entering 0.1 as concentration and 7 as pKa we get pH=4.0 which confirms our calculations.

20 mL of 0.1 M solution of weak acid was mixed with 8  mL 0.1 M solution of NaOH. Measured pH was 5.12. Calculate pKa.

In the case of not-so-weak acids you can assume the neutralization reaction is quantitative (all of the strong base reacts with the weak acid) and the pH of the solution is described by Henderson-Hasselbalch equation:

[pic]15.2

All you have to do is to find out concentrations of A- and HA. However, it is worth of noting here, that you can save part of the calculations, replacing concentrations with numbers of moles - volume is the same for both substances so it cancels out.

Solution was prepared using 20×0.1=2 mmol of acid and 8×0.1=0.8 mmol of strong base. If the reaction was quantitative after the neutralization there was 1.2 mmol of acid HA left and 0.8 mmol of A-. Inserting these values into Henderson-Hasselbalch equation gives 5.12=pKa+log(0.8/1.2). Solving for pKa you get pKa=5.12-log(0.8/1.2)=5.12+0.18=5.30.

pH calculator can help you check this result. Enter 5.3 as pKa, 0.1 as concentration of acid, select sodium hydroxide as a base, enter 0.1 as its concentration, finally select Ca+Cb+Va+Vb as a way of expressing solution composition and enter 20 and 8 as acid and base solutions volumes - and the pH displayed will be 5.12, confirming the result.

Note that for strong or very weak acids and/or for very diluted solutions this method may give wrong results - when pH is further changed by dissociation or hydrolyzis or when the water autodissociation can't be neglected.

Calculate the pH of a 0.2M solution of ammonia. Ka=5.62×10-10.

Ammonia is a weak base, so the most convenient approach is to calculate pOH using Kb, and then to convert it to pH.

From the Brønsted-Lowry theory we know that Ka×Kb=Kw, that allows us to calculate Kb=Kw/Ka=10-14/5.62×10-10=1.78×10-5.

Calculation of the pOH will be now identical with the calculation of pH of a weak acid.

Let's try first with the simplified formula 8.17:

[pic]8.17

If we put concentration and Kb values into formula we get [OH+]=0.00189 M and pOH=2.72. Is it correct? To be sure we have to check if the 5% rule is obeyed, that is if the dissociation fraction is less then 5%.

[pic]1.3

Dissociated base concentration is the same as [OH-] - if we assume that we can neglect water dissociation. It is obvious that we can, as 0.00189 M is about five orders of magnitude larger than the concentration of OH- from the water. So dissociation fraction is 0.00189/0.2 - below 1%, much less then 5%, so the 5% rule is obeyed.

Now we need the final touch - we know pOH=2.72, but we need pH. Time to use equation 1.6. pH=14-2.72=11.28.

Our pH calculator shows 11.27. Difference in the last digit can be probably attributed to different dissociation constant used.

Calculate pH of 0.01M aniline hydrochloride. Aniline pKb=9.4.

Aniline in solution of its hydrochloride is in form of conjugate acid. To calculate pH of such solution we have to find pKa for aniline and treat it like weak acid.

Knowing pKb we can find pKa from the equation 2.5:

[pic]2.5

pKa=14-9.4=4.6 - and this value we will put into simplified formula 8.13.

[pic]8.13

Result is [H+]=5.0×10-4 M and pH=3.30. We have used simplified formula, so now we have to check if the 5% rule is obeyed:

[pic]1.3

Dissociation fraction is 5.0×10-4/0.01 - exactly 5%. We were allowed to use equation 8.13.

Our pH calculator doesn't calculate pH for salts, but salt is nothing else but equimolar solution of acid and base. BATE calculates pH as 3.31 - where is the source of difference?

First suspect is 5% rule - it doesn't state results will be exact, but the error will be acceptable. Let's try to calculate the result using more precise formula 8.10.

[pic]8.10

This time [H+]=4.89×10-4 and pH=3.31. This result is more precise - but is it better and worth additional calculations?

No!

Dissociation constant is given only with two significant digits accuracy, so our pH is in fact 3.3 - or even just 3, as the concentration is given with only one significant digit.

pH calculator always calculate the result with full precision - that's what computers are for. It is up to us - humans - to understand the results.

Calculate pH of 0.1M potassium hydrogen oxalate. pKa1=1.25, pKa2=4.27.

This question - if done by hand - requires either approximate method (that gives questionable results) or tedious calculations.

In amphiprotic salt section we have derived equation 12.9

[pic]12.9

According to the analysis we have done (results shown in the table) equation 12.9 gives good results for more concentrated solutions. 0.01 M is a little bit too diluted for the precise result.

PH=1/2(1.25+4.27

) =2.76.

Let's check this result using pH calculator. To create potassium hydrogen oxalate solution you have to select oxalic acid and potassium hydroxide and to enter 0.1 as both concentrations.

pH calculator shows 2.86 - and this is the correct result. Approximate method is very fast - but gives only very rough results.

Calculate pH of solution prepared from 0.1 mole of formic acid and 0.02 mole of NaOH diluted to 1L. pKa=3.75.

This is classic buffer question that we will solve using Henderson-Hasselbalch equation.

[pic]15.2

We need to know concentration of HCOO- and HCOOH. Formic acid has pKa value larger then 2.5 (see pH of buffer section) so we can assume stoichiometry is solely responsible for concentrations of both acid and conjugate base. If so, after neutralization has taken place [HCOOH]=0.1-0.02=0.08 M and [HCOO-]=0.02 M. Putting these values into Henderson-Hasselbalch equation we get:

pH=3.75+log(0.02/0.08)=3.15.

Using pH calculator we get 3.17. Our rule of thumb (the one about pKa>=2.5) states that difference between real pH and calculated pH will be acceptable, not zero.

You have 100 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.5?

Adding hydrochloric acid to the solution of ammonia (base) we create a conjugate acid NH4+. Ratio of their concentrations at pH 9.5 will be given by the Henderson-Hasselbalch equation.

[pic]15.2

If all added acid reacts with ammonia, amounts of conjugate acid NH4+ created and base NH3 left is defined by the stoichiometry of the neutralization reaction:

[pic]qbq2.1

and

[pic]qbq2.2

[pic]qbq2.3

qbq2.3 is equivalent to

[pic]qbq2.4

where na and nb are number of moles of acid and conjugate base, nNH3start is number of moles of ammonia at the beginning (0.1 mole - n=CV) and nHCl is number of moles of acid added. We are using number of moles as it makes calculation much easier - this way we don't have to calculate how the concentrations change due to the dilution once the acid is added to original ammonia solution. We can do that thanks to the fact that in the Henderson-Hasselbalch equation volumes can be canceled out and replaced by mole ratio.

Putting pH=9.5 into the Henderson-Hasselbalch equation we get

[pic]qbq2.5

Finally we have a set of equations qbq2.4 and qbq2.5. When solved they give na=0.036 and nb=0.064. From qbq2.2 we get that we have to add 0.036 mole of HCl, or 36 mL of 1 M solution.

We can use pH calculator to check the result. Create a solution containing both ammonia and hydrochloric acid. Enter 1 as both concentrations. From the drop down list above pH sign select Ca+Cb+Va+Vb. Enter 36 as acid volume and 100 as base volume. Perfect!

Note: if you need program that will help in buffer calculation, our pH calculator is not suited for the task, however, you can try our Buffer Maker - the buffer calculator. Buffer calculator was specifically designed to help in buffer preparation, so - given pH - it will automatically calculate recipe of any buffer you may need.

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