posted by Fai on .
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?
Please helps me to solve it for me
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
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.
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.
I do not still understand how to calculate PH 1.9 = 0.0126M
I want step by step to calculate for me.
Thank you for your help.