# PH-calculation question.

1. Nov 23, 2004

### Mathman23

I have calculated the PH-value for this following solution:

100 mL 0,102 M HCL and 100 mL 0,0780 M NaOH.

To calculate the pH in the this solution first I must calculate the number of moles $n_{[H_3O^+]}$.

$n_{[H_3 O^{+}]} = 0,100 L \cdot 0,0102 \ mol/L + 0,100 L \cdot 0,0780 mol/L = 0,018 mol$

This means that $[H_3 O^{+}] = \frac{0,018 mol}{0,200 L} = 0,09 mol/L$

pH for the solution is then $pH = \textrm{-log}(0,09) = 1,05$

I would appriciate if somebody would look at my calculation and then tell me if its accurate ??

Many thanks in advance.

Sincerely

Fred

2. Nov 23, 2004

### gravenewworld

HCl+NaOH------> NaCl(aq) + H2O

you need 1 mole of NaOH to neutralize 1 mole of HCl according to the balanced equation.

So you have .102 M HCl which means you have .102 moles HCl/ 1 L of solution so if you have 100 mL of .102 M HCL you have .102 moles HCl/1 L soln x 1L soln/1000mL x 100mL= .0102 moles of HCl. Doing the same thing for NaOH, you have .0078 moles of NaOH. Since NaOH and HCl react in a 1:1 ratio you can obviously see that NaOH is the limiting reagent. Thus .0102moles HCl-.0078 moles HCl =.0024 moles of HCl left. Now I am assuming that you are supposed to assume that 100 mL of HCl+100mL of NaOH=200 mL solution to make things easier, but in reality volumes don't add. Anyway assuming that they do, you have .0024 moles HCl/200mL solution x 1000mL/1L =.012 M HCl so pH=-log[H+]=-log[.012]=1.92

3. Nov 23, 2004

### gravenewworld

Also, HCl is a strong acid so it will dissociate completely

Last edited: Nov 24, 2004
4. Nov 24, 2004

### dustwave

HCl + NaOH --> Na+(aq) + Cl-(aq) + H2O

Na and Cl wont react and form NaCl because the corresponding base of a strong acid is weak and the corresponding acid of a base is also weak, so the formula for those who react is:

H+ + OH- --> H20

5. Nov 25, 2004

### gravenewworld

Thats why I wrote NaCl (aq). That implies that the Na and Cl split apart into ions.