Question

A student has a 1.0 L solution of 2.0 M HCl and wants to increase the HCl concentration to 3.0 M. Which action will produce the intended outcome?
	A) Add 1.0 L of 4.0 M HCl to the existing solution
	B) Add 1.0 L of 1.0 M HCl to the existing solution
	C) Add 1.0 L of 3.0 M NaCl to the existing solution
	D) Add 1.0 L of 12 M HCl to the existing solution
	E) Add 1.0 L of water to the existing solution

Answer 1

The 1 L of 2M HCL plus the addition of the additional substance will tamper the strength of the HCL and this can be calculated using the equation

V1* S1+V2*S2=(V1+V2)*S

where S1 ans S2 are the strengths of the individual solutions ,V1 and V2 are the volumes of the individual solutions and the S is the final strength of the solution.

For option A,

1*2.0+1*4.0=S*2

S=3.0

Therefore the strength of the resulting solution is 3.0 M,hence option A is correct.

All the other options involving the addition of acid is done the same way as shown above.

All the other options are wrong and somehow irrelevant to the present discussion.For eg. the dilution of the water will only decrease the strength of the HCL present and thus E is ruled out.

Option C:- The addition of NaCl increases the concentration of Cl- to 3M (from NaCl ) + 2.0 M (from HCl) thus the Cl- concentration inceases dramatically. But the H+ concentration remains the same with the increasing volume of the solution and the strength decreases from 2.0M to (1*2.0)/2 =1.0 M.