How many grams of NaCN would you need to dissolve in enough water to make exactl

Question

Explanation / Answer

First of all, you need the appropriate chemical equation: NaCN + H2O --> HCN + OH- Now, you can set up a chart NaCN + H2O --> HCN + OH- initial: ?M 0M 0M change: -x +x +x equilibrium: (?-x)M xM xM We need to find the equilibrium concentration of NaCN. We know that the pH is 10.00, so we can use this to find pOH. pH + pOH = 14.00 pOH = 14.00 - 10.00 pOH = 4.00 Now that you know pOH, you can find the equilibrium concentration of OH-. pOH= -log[OH-] 4.00 = -log[OH-] -4.00 = log[OH-] [OH-] = 1.00E-4 You can now fill this value in for the chart above. NaCN + H2O --> HCN + OH- initial: ?M 0M 0M change: -1.00E-4M +1.00E-4M +1.00E-4 equilibrium: (?-1.00E-4)M 1.00E-4M 1.00E-4M Remember, we needed to find the equilibrium concentration of NaCN. At equilibrium we see that the concentraion is the initial concentration of NaCN - 1.00E-4 The Kb value (you should be able to look this up in a chart in your book) is equal to 2.53E-5. Kb = ( [HCN][OH-] ) / [NaCN] We can now substitute in the equilibrium values we have found. This will allow us to find the unknown initial concentration of NaCN. 2.53E-5 = [1.00E-4][1.00E-4] / [x -1.00E-4] Cross multiply to get: (2.53E-5)x = 1.00E-8 x = 3.95E-4 So, the initial concentration of NaCN was 3.95E-4. Remember that at equilibrium the concentraion of NaCN is the initial concentration of NaCN - 1.00E-4. So, the equilibrium concentration of NaCN = 3.95E-4 + 1.00E-4 = 4.95E-4 Finally, we can find how many grams of NaCN we need. (The molar mass of NaCN is 49.0 g/mol, and 1 L = 1000 mL) (4.95E-4 mol NaCN/ 1L) x (49.0 g NaCN/ 1mol NaCN) x(1 L/1000 mL) x (450 mL) = 0.0109147 g NaCN

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.