Chemistry; acid base help!!!!?

Question

I know how to do most of acid base questions but i'm having trouble with these; in the problems use pH=-log(H+)

200mL of a 0.1mol/L HCl solution is added to 198mL of a 0.1mol/L NaOH solution. Calculate the hydrogen ion and hydroxide ion concentrations in the resulting solution.

0.25g of sodium hydroxide are added to 115mL of a nitric acid solution of pH 1.50. Determine the pH of the resulting solution. Assume volume does not change on adding the solid.

10.6g of anhydrous sodium carbonate are dissolved in 560mL of a 0.54mol/L HCl solution
a) Calculate the pH of the final solution, assuming volume does not change on adding the solid
b) What volume of water must be added to give a solution having pH 1

Any help is great, thx

Trevor H · Accepted Answer

Question 1: 
 You know that if you react NaOH with HCl you produce NaCl. This is a neutral salt and it does not affect the pH of the final solution. 
You also know that HCl reacts with NaOH on 1:1 molar basis. 
Therefore 198mL of 0.1M NaOH solution will react with 198mL of 0.1M HCl. Because you have 200mL of the HCl solution, you will have 2mL of the 0.1M HCl unreacted. The final volume of the solution will be 200+198 = 398mL of solution

Now what is the molarity of the HCl in this solution: Use the dilution equation:
M1V1 = M2V2
M1-398 = 0.1*2
M1 =  0.2/398
M1 = 5.025*10^-4 M HCl solution

HCl is a strong acid that dissociates completely:
Therefore in a 5.025*10^-4M solution you have:
 [H+] = 5.025*10^-4M 
And:
 [H+] [OH-] = 10^-14
[OH-] = 10^-14 / ( 5.025*10^-4)
[OH-] = 1.99*10^-11M 

I am not going to do the other questions for you - I have shown you how to go about these problems. If you really have a problem email me, but I would like you to try yourself, and if you cannot make progress, send me your working.

blackbeauty14 · Answer

In these types of problems, all you need to do is to calculate the no. of moles of the acid and basic species in the amount of (meaning volume) solution you have.

so 200 ml of 0.1 M HCL will have 0.2 L*0.1 M = 0.02 moles of replaceable H+
198 ml of 0.1 M NaOH will have 0.1*0.198L = 0.0198 moles of OH- ions

Since the OH- concentration is lowered, they will be completely neutralized by the excess H+, and that will leave behind 0.02 -0.0198 = 0.002 moles of H+ ions free in 398 mL (i.e., 200 + 198 ml) of solution

For the second question, using the equation you mentioned (pH = -log H+) you first calculate the H+ concentration in the 115 ml of nitric acid solution.  Then calculate the number of moles of OH- present in 0.25 g NaOH (basically equal to 0.25/40).  Again you use the same approach as above, part of the H+ will get neutralized.  You take the remaining H+ concentration and insert in the same equation, and then take a negative log to get the pH of the resulting solution,

Same approach as above for the third part.  Unfortunately, I do not have a calculator handy to do the logs.  I am absolutely confident that you can do this.

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Chemistry; acid base help!!!!?

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