What mass of solid NaCN must be added to 500 mL of 0.50 M HCN to prepare a buffe

Question

What mass of solid NaCN must be added to 500 mL of 0.50 M HCN to prepare a buffer solution with a pH of 9.50? pKa for HCN is 9.21 For the following equilibrium: CaC_2O_4(s) doubleheadarrow Ca^2+_(aq) + C_2O_4^2- (aq) If the concentration of calcium ions is 0.036M and oxalate is 2.4 times 10^-7 M, will a precipitate occur? Why, why not? K_ap = 2.3 times 10^-9 for calcium oxalate If the concentration of calcium ions is 6.98 times 10^-8 M and oxalate is 1.3 times 10^-4 M, will a precipitate occur? Why, why not? K_ap = 2.3 times 10^-9 for calcium oxalate

Explanation / Answer

5)

pH = pKa + log {[NaCN]/[HCN]}

9.50 = 9.21 + log {[NaCN]/0.50}

log {[NaCN]/0.50} = 0.29

[NaCN]/0.50 = 1.95

[NaCN] = 0.975 M

molar mass of NaCN = 49 g/mol

[NaCN] = mass / (molar mass * volume in L)

0.975 = mass / (49*0.5 )

mass = 23.9 g

Answer: 23.9 g

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