0.00735 grams of sodium cyanide would you need to dissolve in enough water to make exactly 250 ml of solution with a ph of 10.00.
pOH = 4.00
[OH-] = 10^-4 M = [HCN]
2.0 x 10^-5 = (10^-4)^2 / X
X = 0.000500 M
initial concentration NaCN = 0.000500 + 10^-4 =0.000600 M
moles = 0.250 L x 0.000600=0.00015
mass = 0.00015 x 49 g/mol=0.00735 g
pOH = 4.00
[OH-] = 10^-4 M = [HCN]
2.0 x 10^-5 = (10^-4)^2 / X
X = 0.000500 M
initial concentration NaCN = 0.000500 + 10^-4 =0.000600 M
moles = 0.250 L x 0.000600=0.00015
mass = 0.00015 x 49 g/mol=0.00735 g
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