what approx volume of 0.4. M Ba(OH)2 must be added to 50.0ml of 0.30 M NaOH to get a solution in which the molarity of the OH- ions is 0.50M?
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27
Let us suppose that x mL of 0.40 M Ba(OH)2 soon is required to be added.
Assuming that Ba(OH)2 dissociates completely, number of millimoles of OH- in solution = x *0.80 mmol
(Since each formula unit of Ba(OH)2 gives 2 OH- ions.)
Number of millimoles of OH- due to dissociation of NaOH = 50*0.30 =15 mmol
Total number of millimoles of OH- in solution= 15+x*0.80 mmol
Total number of millimoles of OH- required for 0.50 M solution = (50+x)mL*0.50 mmol/mL
=25 +x*0.50 mmol
Hence, 15+0.80*x =25 +x*0.50
=> 0.30*x =10
x =10/0.30 =33.33
Approximately 33 mL of the Ba(OH)2 solution needs to be added.
Assuming that Ba(OH)2 dissociates completely, number of millimoles of OH- in solution = x *0.80 mmol
(Since each formula unit of Ba(OH)2 gives 2 OH- ions.)
Number of millimoles of OH- due to dissociation of NaOH = 50*0.30 =15 mmol
Total number of millimoles of OH- in solution= 15+x*0.80 mmol
Total number of millimoles of OH- required for 0.50 M solution = (50+x)mL*0.50 mmol/mL
=25 +x*0.50 mmol
Hence, 15+0.80*x =25 +x*0.50
=> 0.30*x =10
x =10/0.30 =33.33
Approximately 33 mL of the Ba(OH)2 solution needs to be added.
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1
Answer:
Explanation:
N1V1+N2V2=Ntotal*Vtotal
0.4*2*V+ 0.30*1*50 = 0.15(50+v)
0.2v + 15 = 0.5*50 + 0.5v
0.3v = 10
V = 10÷0.3
V = 33
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