- Thread starter Helly
- Start date

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

B. [H] = 15 . 10^-4

C. [H] = 5 . 10^-4

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

Looking at your figures

(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

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 ?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.

Looking at your figures

(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

Sodecreasingorder 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.

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.I think my [H] still wrong.

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

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?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

I think the key answer is wrong, ofc more HCl left makes lower pH, condition C is less acid than B right?

Last edited:

What is this meant to mean ?For B , NaCO2 doesnt pH?

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?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?

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

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

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.For B, HCl all reacted with Na2CO3? Means only left Na2CO3?

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 ?

For B, HCl all reacted with Na2CO3? Means only left Na2CO3?

The pH for B is netral?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 ?

Similar Chemistry Discussions | Chemistry Forum | Date |
---|---|---|

Finding PH | High School Chemistry | |

Finding solubility constants in a titration | High School Chemistry | |

Finding pH of solution (KOH, HCl, NaCl) | High School Chemistry | |

Finding unknown element | Undergraduate Chemistry |