# chemistry

Calculate the pH of 0.3 M HCl solution. Calculate the pH when 100 mL of a 0.5 M solution of NaOH are added to 300 mL of the HCl solution.

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1. pH = -log (H^+) = -log(HCl)

.......HCl + NaOH ==> NaCl + H2O
millimols HCl = mL x M = approx 90
mmols NaOH = 50
Excess HCl = 90-50= 40
total volume = 400 mL
M = mmols/mL = 40/400 = 0.1
pH = -log(H^+) = ?

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2. Why and How you took 90 as HCl millimos? Is there any possibility to use ICE table?

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3. Moles are what reacts in chemistry. mols = molarity x liters. I prefer to work in millimols so mmols HCl = mL x M = ?
The problems tells you M HCl is 0.3 and the mL = 300 so 300 x 0.3 = 90 mmols. Yes, you can use an ICE table and I used an abbreviated one but here is the full version in use. You really need to read these responses with more care because I've done this problem in more detail than should be required.
mmols HCl = mL x M = 300 x 0.3 = 90
mmols NaOH = 100 x 0.5 = 50
......HCl + NaOH ==> NaCl + H2O
I.....90.....0........0......0
C....-50...-50........50.......
E.....40.....0........50.......

So you have an excess of 40 mmols HCl in a total of 300+100 = 400 mL of solution.
M = mmols/mL = 40/400 = 0.1
Then convert 0.1M HCl to pH.
Basically I've reworked the problem but added three lines for the ICE table.

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