In: Chemistry
1)
Which of the following solutions is a good buffer system?
Which of the following solutions is a good buffer system?
A solution that is 0.10 M HCN and 0.10 M LiCN | |||||||||||||||||||||
A solution that is 0.10 M NaCl and 0.10 M HCl | |||||||||||||||||||||
A solution that is 0.10 M KOH and 0.10 M HNO3 | |||||||||||||||||||||
A solution that is 0.10 M HNO3 and 0.10 M KNO3 | |||||||||||||||||||||
A solution that is 0.10 M HCN and 0.10 M Na Cl 2) Calculate the pH of the solution that results from each of the following mixtures. Part A 55.0 mL of 0.17 M HCHO2 with 70.0 mL of 0.13 M NaCHO2 Express your answer using two decimal places. Part B 115.0 mL of 0.11 M NH3 with 250.0 mL of 0.11 M NH4Cl Express your answer using two decimal places. 3) A 1.0-L buffer solution is 0.125 M in HNO2 and 0.180 M in NaNO2. Part A Determine the concentrations of HNO2 and NaNO2 after addition of 1.5 g HCl. Express your answers using three significant figures separated by a comma. Part B Determine the concentrations of HNO2 and NaNO2 after addition of 1.5 g NaOH. Express your answers using three significant figures separated by a comma. Part C Determine the concentrations of HNO2 and NaNO2 after addition of 1.5 g HI. Express your answers using three significant figures separated by a comma. 4) A ligand is a molecule or ion that acts as a A ligand is a molecule or ion that acts as a
|
1. answer: A: A solution that is 0.10 M HCN and 0.10 M LiCN
In order to search a effitien Buffer sistem, we have to identify (in this case) the weak acid, and his respective conjugate base, as you can see all other options are stong acids. Why not strong acids? it is because for the definition of strong acid, this are completly dissociate in water, for this aspect is that we always search to the weak acir (or base)
2. a)
Finding moles:
moles HCOOH = 0.0550 L x 0.17 M=0.00935
moles HCOO- = 0.0700 L x 0.13 M=0.0091
total volume = 55.0 + 70.0=125 mL = 0.125 L
concentration HCOOH = 0.00935 / 0.125 = 0.0748 M
concentration HCOO- = 0.0091 / 0.125 =0.0728 M
pKa = 3.74
applying Henderson-Hasselbalch equation:
pH = 3.74 + log 0.0728/ 0.0748 =3.73
B)
moles NH3 = 0.115 L x 0.11 M = 0.01265
moles NH4+ = 0.250 L x 0.11 M=0.0275
total volume = (115 + 250)mL=365 mL = 0.365 mL
concentration NH3 = 0.01265 / 0.365 =0.0346 M
concentration NH4+ = 0.0275 / 0.365 =0.0753 M
pOH = 4.74 + log 0.0753/ 0.0346 =5.07
pH = 14 - 5.07 = 8.93