Eboni asked in Science & MathematicsChemistry · 1 decade ago

# Calculate the pH of the following strong acid solutions:5.00mL of 1.00 M HCl diluted to .750L?

Update:

Thanks so much people!!! I had it all the way up to .005 part!! Gosh I love yahoO!!

### 3 Answers

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• 1 decade ago
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5.00 mL x 1.00 M = 5 millimoles of HCl ( 0.005 moles)

0.005 moles HCl / 0.750 L = 0.00667 M

pH = -log(0.00667) = 2.18

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• Anonymous
1 decade ago

1.00 M = 1 mole per liter

5 mL = 0.005 L so there is 0.005 moles of the stuff

now you have 0.005 mole in .750 liters so you divide:

0.005/.750 and you get .001/.15 or (1/150) M thats the new concentration

now because this is a strong acid the hydrogen completely dissociates and you can calculate the pH with this equation:

pH = -log(H+) = -log(1/150) = 2.18 thats your answer please dont drink this solution you already have enough HCl in your stomach!

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• jon
Lv 4
1 decade ago

First find the concentration of HCl after diluting the original solution:

No of moles of HCl = 1.00 * 5 / 1000 = 0.005

0.005 moles in 0.75L = 0.005 * 1 / 0.75 = 0.00667 moles/litre

So, [HCl] = 0.00667 M

Since HCl is a strong acid (completely dissociated into ions),

HCl ---> H^+ + Cl^-

Then [H^+] = 0.00667 M

pH = -log[H^+] = - log(0.00667) = 2.18

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