When 100 ml of 0.1 M KNO 3 , 400 ml of 0.2M HCl and 500 ml of 0.3 M H 2 SO 4 are mixed. Then in the resulting solution
= 0.15 M
= 0.08 M and Cl – = 0.01 M
Text Solution
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(a, b, c, d)
When the three solutions are mixed then the molarity of each solution changes but number of moles does not change. Hence it is similar to dilution.
For KNO 3 M 1 V 1 = M 2 V 2
100 × 0.1 = M 2 × 1000 ⇒ M 2 = 0.01
⇒ resulting molarity of KCl = 0.01
⇒ K + = 0.01 Μ Cl – = 0.01 M
For HCl M 1 V 1 = M 2 V 2
0.2 × 400 = M 2 × 1000 ⇒ M 2 = 0.08
⇒ H + = 0.08 M
= 0.08 M
For H 2 SO 4 M 1 V 1 = M 2 V 2
0.3 × 500 = M 2 × 1000 ⇒ M 2 = 0.15 ⇒ H + = 0.3 M 
= 0.15 M
Hence in the solution K + = 0.01 M; Cl – = 0.01 M; 
= 0.08 M; H + = 0.38 M; 
= 0.15 M
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