# Thread: Ph calculations help!!!!! :(

1. ## Ph calculations help!!!!! :(

1. 1.0ml of 10mol hcl acid are diluted to 1L with distilled water. 250ml of this solution are then further diluted to 1L with distiiled water. Calculate the ph of the final solution

2. 500ml of 0.15 mol sodium hydroxide solution are mixed with 500ml of 0.1mol hcl acid solution. Calculate the ph of the final mixture.

Thanks!!

2. ## Re: Ph calculations help!!!!! :(

I won't give you the answer, but tell you what to do step by step.

First of all, you must remember that the first thing you need to do for any Chemistry calculation is to translate the question into a chemical equation (or sometimes two).

Also, I have realised that your questions are a little incorrect. For the first one, is the concentration supposed to be 10 mol/L? The second question is also incorrect in that sense.

1. Hydrochloric acid is a strong acid meaning that it completely ionises in water. The first thing you must do is write the equation for it ionising in water. This way you get the total molar ratio of acid to hydrogen ions in solution.

Now you need to calculate the moles of acid used, then using the ratio of moles in the balanced equation of complete dissociation you find how many moles of hydrogen ions are in solution.

Then find the concentration of the hydrogen ions in the water.

After that, you use the formula found on the data table for finding pH from Hydrogen ion concentration.

Tell me how you go. Do you need a full solution?

2. For this one, both the acid and base are strong and will completely dissociate in solution.

Now, you need to write the equation for the reaction between Sodium Hydroxide and Hydrochloric acid.

Now, you need to calculate from the data given how many moles of each is present. Then you need to subtract the common amount of moles.

In the end, you will remain with an extra amount of one substance (either the acid or base). Then you use this value of excess moles of one substance to calculate the concentration of excess Hydroxide or Hydrogen ions (whichever is in excess). Then substitute into the formula straight away (if Hydrogen ions are excess) or if Hydroxide ions are in excess you have to use the fact that:

Kw = 10^(-14)

If you need the full solution then I can post it up. But this way you will learn more.

3. ## Re: Ph calculations help!!!!! :(

Thanks for ur help! But i still dont understand the first question did u get 1.6 for the second q?

4. ## Re: Ph calculations help!!!!! :(

For question 1, is it supposed to be 10 moles of Hydrogen Chloride is added OR 10 moles per litre concentration of Hydrochloric acid added?

For question 2, I got 2.3. Though I rushed it a bit and I could be a little wrong. Check this one with your teacher/answers maybe?

(Sorry I don't have a calculator on me at the moment).

mol L^-1

6. ## Re: Ph calculations help!!!!! :(

Originally Posted by littlesmallworld
1. 1.0ml of 10mol/L hcl acid are diluted to 1L with distilled water. 250ml of this solution are then further diluted to 1L with distiiled water. Calculate the ph of the final solution
Thanks!!
First things first:

$HCl_{aq} \rightarrow H^{+}_{(aq)} + Cl^{-}_{(aq)}$

By using the fact that HCl is a strong acid and completely dissociates in water, and that the molar ratio from the balanced equation above for HCl to Hydrogen ions is 1:1, then we know that the moles of HCl dissociated is equal to the moles of Hydrogen ions produced. So hence:

$n_{HCl} = n_{H^{+}}$

Calculate the new concentration of the Hydrogen ions in solution:

$n = c_{1} v_{1} = c_{2} v_{2} \\ \\ \\ c_{2} = \frac{c_{1} v{1} }{ v_{2} } \\ \\ \\ c_{2} = \frac{ (10 mol \ L^{-1} ) \left ( \frac{1}{1000} \ L \right ) }{ 1 \ L} \\ \\ \\ c_{2} = 10^{-2} \ mol \ L^{-1}$

Now we know the concentration of the Hydrogen ions in solution, we can just substitute into the following formula found on the data sheet to find the final pH of solution:

$pH = - log_{10} [H^{+}] \\ \\ \\ pH = - log_{10} [10^{-2}] \\ \\ \\ \therefore pH = 2$

Originally Posted by littlesmallworld
2. 500ml of 0.15 mol sodium hydroxide solution are mixed with 500ml of 0.1mol hcl acid solution. Calculate the ph of the final mixture.

Did you get the correct solution for question 2?

Did you check the answer with someone else?

7. ## Re: Ph calculations help!!!!! :(

Originally Posted by anomalousdecay
First things first:

$HCl_{aq} \rightarrow H^{+}_{(aq)} + Cl^{-}_{(aq)}$

By using the fact that HCl is a strong acid and completely dissociates in water, and that the molar ratio from the balanced equation above for HCl to Hydrogen ions is 1:1, then we know that the moles of HCl dissociated is equal to the moles of Hydrogen ions produced. So hence:

$n_{HCl} = n_{H^{+}}$

Calculate the new concentration of the Hydrogen ions in solution:
is q2 1.6?

$n = c_{1} v_{1} = c_{2} v_{2} \\ \\ \\ c_{2} = \frac{c_{1} v{1} }{ v_{2} } \\ \\ \\ c_{2} = \frac{ (10 mol \ L^{-1} ) \left ( \frac{1}{1000} \ L \right ) }{ 1 \ L} \\ \\ \\ c_{2} = 10^{-2} \ mol \ L^{-1}$

Now we know the concentration of the Hydrogen ions in solution, we can just substitute into the following formula found on the data sheet to find the final pH of solution:

$pH = - log_{10} [H^{+}] \\ \\ \\ pH = - log_{10} [10^{-2}] \\ \\ \\ \therefore pH = 2$

Did you get the correct solution for question 2?

Did you check the answer with someone else?

so is q2 1.6??

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