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1. A. 1 xx is the hydrogen ion concentration.
B. pH of the solution is 10
C. The solution is basic.
2. The molarity of NaCl in 2.8 litres of water is 1.18 M
3. 1.64 M is the molarity of new solution.
4. 0.64 M is the molarity of the acid or HCl used.
Explanation:
1. Data given [OH-] =1 x 10-4 mol/L.
A) = [
O+] [OH]-
= 1×
[O+]= 1×
÷ 1 x 10-4
= 1 xx is the hydrogen ion concentration.
B) pH =-log [O+]
pH = -log[-1x ]
pH = 10
c) pH value of 10 indicates that it is a basic solution because it is greater than 7 and on pH scale more than 7 value indicates basicity.
2. Molarity is calculated by the formula
Molarity =
Number of moles can be calculated as:
n =
n = ( atomic weight of NaCl is 58.44 gm/mole)
n= 3.33 moles
Now the molarity of NaCl in 2.8 litres of water is calculated as:
Molarity =
M =
M = 1.18
the molarity of NaCl in 2.8 litres of water is 1.18 M
3. Data given:
Initial volume of the HCl solution V1 = 152 ml, initial molarity M1 = 3
final volume of the diluted HCl V2= 750 ml, final molarity M2= Unknown
The initial and final volumes are converted into litres in calculation.
So, M1V1 = M2V2
3× = M2 ×
M2 =
M2 = 1.64 M is the molarity of new solution.
4. In titration the formula used is: M acid x V acid = M base x V base
Volume of base V1= 20 ml Molarity of base M1 = 0.24 M
volume of the acid V2 = 31 ml Molarity of the acid = unknown
the volume will be converted to litres.
So applying the formula,
M acid x V acid = M base x V base
0.24 x = Macid x
0.0048 = Macid x 0.031
Macid=
M base= 0.64 M is the molarity of the acid or HCl used.