a. What is the pH of a buffer obtained by adding 9.8grams of NaCN (MW=49 g/mole) to 100ml of .20 M HCN?

Consider HCN (Ka= 5.8 x 10^-10) and NaCN

b. What is the pH of this buffer after .01 moles of OH- are added to this solution?

Solutions

Expert Solution

no of moles of NaCN = W/G.M.Wt

= 9.8/49 = 0.2moles

no od moles of HCN = molarity * volume in L

= 0.2*0.1 = 0.02 moles

PKa = -logKa

= -log5.8*10^-10 = 9.2365

PH = PKa +log[NaCN]/HCN]

= 9.2365 +log0.2/0.02

= 9.2365 +1

= 10.2365

no of moles of NaCN after addition 0.01moles of OH^- = 0.2+0.01 = 0.21moles

no of moles of HCn after addition of 0.01 moles of OH^- = 0.02-0.01 = 0.01moles

PH = PKa + log[NaCN]/[HCN]

= 9.2365+log0.21/0.01

= 9.2365+1.3222 =10.5587

queen_honey_blossom answered 1 year ago

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