# Calculate the pH of the resulting solution if 33.0 mL of 0.330 M HCl(aq) added to?

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(a) 43.0 mL of 0.330 M NaOH(aq).

(b) 23.0 mL of 0.430 M NaOH(aq).

i understand most of my chem but im really lost on this acid and base stuff thanks for any help you can give

• HCl reacts with NaOH according to the equation:

HCl + NaOH → NaCl + H2O

1mol HCl reacts with 1 mol NaOH

a) mol HCl in 33.0mL of 0.330M solution = 33/1000*0.33 0.0109 mol HCl

Mol NaOH in 43.0mL of 0.330M solution = 43/1000*0.33 0.0142 mol NaOH

These will react to produce 0.0109 mol NaCl and 0.0142 – 0.0109 = 0.0033 mol NaOH unreacted

You ave 0.0033mol NaOH dissolved in 43+33 = 76mL = 0.076L solution

Molarity of NaOH solution = 0.0033/0.076 = 0.0433M solution

If [NaOH]= 0.0433M , then [H- = 0.0433M

To determine pH you require [H+]

[H+] [OH-] = 10^-14

[H+] = 10^-14/ [OH-]

[H+] = 2.31*10^-13

pH = -log [H+]

pH = -log (2.31*10^-13)

pH = 12.6

b) You are still reacting 0.0109 ml HCl

Mol NaOH in 23.0mL of 0.430M solution = 23/1000*0.430 = 0.00989 mol NaOH

These will react to produce 0.00989 mol NaCl and there will be 0.0109- 0.00989 = 0.00101 mol HCl remaining unreacted. This is dissolved in 33+23 = 56mL= 0.056L solution

Molarity of HCl solution = 0.00101/0.056 = 0.0180M solution

If [HCl ] = 0.0180, then [H+] = 0.0180M

pH = -log [H+]

pH = -log 0.0180

pH = 1.74

You understand that the NaCl produced by the reaction has no effect on the pH of the solution.

• 1) Find the number of moles of hydrogen that are produced from HCl and the Number of Moles of Hydroxide from NaOH using their respective volumes and molarities.

2)Since we have a strong acid and a strong base we do not need to worry about equilibrium therefore when mixed, they will both react to either (a) neutralization, (b) some hydrogen will remain, (c) some hydroxide will remain. To calculate this just subtract the moles of hydrogen from the moles of hydroxide or the moles of hydroxide from the moles of hydrogen (which ever is bigger first).

(a) if you get 0 it means you have neutralized therfore pH = 7

(b) if your hydrogen moles was bigger you will have excess hydrogen and therfore be acidic

calculate pH using pH= -log( [H] )

(c) If hydroxide moles was bigger you will have excess hydroxide and therfore be basic

calculate pOH using pOH = – log( [OH] ) then convert to ph using pH = 14 – pOH.

To work out (a) as an example:

Since it is a solution of 33.0 mL of .330 M HCl i know that i have .0109 mols H, using the same process for NaOH i get .0185 mols OH. Since i have more Hydroxide ions than hydrogen ions i know my solution will be basic therfore after the reacion i will have .0185 – .0109 = .00760 Hydroxide ions

left. pOH then equals – log( .00760 ) = 2.12 then i can get pH using pH = 14 – 2.12 which yields a final pH of 11.88.

I hope the helps and makes some sense 🙂

• why no longer upload a established indicator to the answer and then do the titration. wait, are you doing this in genuine existence? is there a formula for pH? i do not imagine so. the prevalent indicator will provide a colour, then there is this peice of paper that has the size and proper rang of colors, positioned it next to the try tube and tournament it up.

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