# chemistry

A solution of sodium cyanide NACN has a PH of 12.10. How many grams of NACN are in 425ml of solution with the same PH?

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1. How to calculate PH 1.9 = 0.0126M Why?
How to calculate PH 1.8 = ?
How to calculate PH 20 = ?
How to calculate PH 21 = ?

Pls helps me

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2. pH = -log(H^+)
1.9 = -log(H^+)
-1.9 = log(H^+).
(H^+) = 0.01259 which rounds to 0.126
The others are done the same way.

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3. pH = 12.10
pH + pOH = pKw = 14
Substitute and sole for pOH. You should get 1.9 for pOH, then
pOH = -log(OH^-) and (OH^-) = 0.0126M.

0.0126M mans 0.0126 mols/L soln. For 425 mL you will have 0.0126 mol x (425/1000) = about 0.00536 mols. Then grams = mols x molar mass NaCN.

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4. I do not still understand how to calculate PH 1.9 = 0.0126M
I want step by step to calculate for me.

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