# Finding pH

#### studiot

A. [H] = 10^-3
B. [H] = 15 . 10^-4
C. [H] = 5 . 10^-4
Well to start with do you think there is any precipitate with the stated mixtures?

The answer is no, sodium salts are generally very soluble.

(assuming your hydrogen ion concentrations are accurate) 15 x10^-4 = 1.5 x10^-3 so A and and B are the wrong way round in your list.

So
A : pH = -log[.001] = 3

B : pH = -log[.0015] = 2.8

C : pH = -log[.0005] = 3.3

So decreasing order would be C > B > A

Remember that decreasing order means increasing [H} concentration ie increasing acidity or decreasing alkalinity.

Note I haven't checked your figures for [H] yet.

• topsquark

#### Helly

Well to start with do you think there is any precipitate with the stated mixtures?

The answer is no, sodium salts are generally very soluble.

(assuming your hydrogen ion concentrations are accurate) 15 x10^-4 = 1.5 x10^-3 so A and and B are the wrong way round in your list.

So
A : pH = -log[.001] = 3

B : pH = -log[.0015] = 2.8

C : pH = -log[.0005] = 3.3

So decreasing order would be C > B > A

Remember that decreasing order means increasing [H} concentration ie increasing acidity or decreasing alkalinity.

Note I haven't checked your figures for [H] yet.
I think my [H] still wrong. The answer is B>C>A. How to find [H] for them ?

#### studiot

I think my [H] still wrong.
So do I so I will do the first one for you and then you can see how myou get on with B ans C.

$$\displaystyle {H_2}S{O_4} \to 2{H^ + } + SO_4^{2 - }$$

So this says that 15mL of 0.1M sulphuric acid ionises into 15 mL containing 0.2M hydrogen ions and 0.1M sulphate ions.
The sulphate ions remain in solution so we do not need to consider them further.

15mL of 0.2M contains 15/1000 * 0.2 moles of H+ ions = 3 x10^-3 moles.

The sodium hydroxide ionises as follows

$$\displaystyle NaOH \to N{a^ + } + O{H^ - }$$

This says that 10mL of 0.1M hydroxide ionises into 10mL containing 0.1M sodium ions and 0.1M hydroxyl ions.
Again we do not need to consider the sodium ions further, they remain in solution.

But all the hydroxyl ions react with some of the hydrogen ions. Since there are more hydrogen ions there are some left over.

10mL of 0.1M contains 10/1000*0.1 moles of OH- ions = 1 x10^-3 moles.

Take these away from the number of hydrogen ions leaves the net amount of H+ ions

(3 x10^-3) - (1 x10^-3) = 2 x10^-3

But you have also combined the volumes so you now have 15mL + 10mL = 25mL

So there are 2 x10^-3 moles of H+ in 25 mls of combined solution or (2 x10^-3) * (1000/25) moles in 1000 mLs or 1 litre.

So the final concentration of hydrogen ions is [80 x 10^-3 ]M or 0.08M

So the pH is -log(0.08) = 1.1

Compare this with the pH of the dissociated acid = -log(0.2) = 0.7

• Helly and topsquark

#### Helly

So do I so I will do the first one for you and then you can see how myou get on with B ans C.

$$\displaystyle {H_2}S{O_4} \to 2{H^ + } + SO_4^{2 - }$$

So this says that 15mL of 0.1M sulphuric acid ionises into 15 mL containing 0.2M hydrogen ions and 0.1M sulphate ions.
The sulphate ions remain in solution so we do not need to consider them further.

15mL of 0.2M contains 15/1000 * 0.2 moles of H+ ions = 3 x10^-3 moles.

The sodium hydroxide ionises as follows

$$\displaystyle NaOH \to N{a^ + } + O{H^ - }$$

This says that 10mL of 0.1M hydroxide ionises into 10mL containing 0.1M sodium ions and 0.1M hydroxyl ions.
Again we do not need to consider the sodium ions further, they remain in solution.

But all the hydroxyl ions react with some of the hydrogen ions. Since there are more hydrogen ions there are some left over.

10mL of 0.1M contains 10/1000*0.1 moles of OH- ions = 1 x10^-3 moles.

Take these away from the number of hydrogen ions leaves the net amount of H+ ions

(3 x10^-3) - (1 x10^-3) = 2 x10^-3

But you have also combined the volumes so you now have 15mL + 10mL = 25mL

So there are 2 x10^-3 moles of H+ in 25 mls of combined solution or (2 x10^-3) * (1000/25) moles in 1000 mLs or 1 litre.

So the final concentration of hydrogen ions is [80 x 10^-3 ]M or 0.08M

So the pH is -log(0.08) = 1.1

Compare this with the pH of the dissociated acid = -log(0.2) = 0.7
For B , NaCO2 doesnt pH?
I think the key answer is wrong, ofc more HCl left makes lower pH, condition C is less acid than B right?

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#### studiot

For B , NaCO2 doesnt pH?
What is this meant to mean ?
The question says $$\displaystyle N{a_2}C{O_3}$$

Why have you not written out the reaction equations like I showed you?

Is any gas evolved when you neutralise hydrochloric acid with sodium carbonate?

• Helly and topsquark

#### Helly

What is this meant to mean ?
The question says $$\displaystyle N{a_2}C{O_3}$$

Why have you not written out the reaction equations like I showed you?

Is any gas evolved when you neutralise hydrochloric acid with sodium carbonate?
For B, HCl all reacted with Na2CO3? Means only left Na2CO3?
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

#### studiot

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
Yes that's right. That is the overall reaction.

For B, HCl all reacted with Na2CO3? Means only left Na2CO3?
So to answer your second question (and earlier one about pH) you need to break this down into three part reactions, just as in the model I gave for part A.
That will show where the H+ ions come from.

$$\displaystyle N{a_2}C{O_3} \to 2N{a^ + } + CO_3^{2 - }$$

$$\displaystyle 2HCl \to 2{H^ + } + 2C{l^ - }$$

$$\displaystyle 2{H^ + } + CO_3^{2 - } \to {H_2}O + CO_2^ \uparrow$$

So these equations say that 2 moles of hydrochloric acid react with 1 mole of sodium carbonate.

So can you calculate how many moles of hydrochloric acid there are in 15 mL of 0.1M solution
and
How many moles of sodium carbonate there are in 10mL of 0.1M solution

?

It helps to say the words like this to yourself.

When you have done this can you calculate how many moles of hydrochloric acid or sodium carbonate are left over or not used in the reaction if any at all ?

• Helly

#### Helly

For B, HCl all reacted with Na2CO3? Means only left Na2CO3?
Yes that's right. That is the overall reaction.

So to answer your second question (and earlier one about pH) you need to break this down into three part reactions, just as in the model I gave for part A.
That will show where the H+ ions come from.

$$\displaystyle N{a_2}C{O_3} \to 2N{a^ + } + CO_3^{2 - }$$

$$\displaystyle 2HCl \to 2{H^ + } + 2C{l^ - }$$

$$\displaystyle 2{H^ + } + CO_3^{2 - } \to {H_2}O + CO_2^ \uparrow$$

So these equations say that 2 moles of hydrochloric acid react with 1 mole of sodium carbonate.

So can you calculate how many moles of hydrochloric acid there are in 15 mL of 0.1M solution
and
How many moles of sodium carbonate there are in 10mL of 0.1M solution

?

It helps to say the words like this to yourself.

When you have done this can you calculate how many moles of hydrochloric acid or sodium carbonate are left over or not used in the reaction if any at all ?
The pH for B is netral?

#### studiot

No, why should it be?

But I think this is beyond high school chemistry.