It is necessary to make 500 mL of a 0.375 M solution of hydrochloric acid (HCl) using distilled water and full strength hydrochloric acid (M=12.0). what volume of the full strength HCl and what volume of water are necessary to make the desired solution?

15.625 mL

Explanation:

To solve this problem, we have to make use of the dilution formula. By so doing we can obtain the volume of stock solution that should be required to prepare the specified concentration and volume of the dilute acid as shown in the given question.

The dilution formula is;

C1V1= C2V2

Where;

C1= concentration of stock solution = 12.0 M

V1= volume of stock solution = ???

C2= concentration of dilute solution = 0.375 M

V2= volume of dilute solution= 500 mL

Making V1 the subject of the formula;

V1= C2V2/C1

V1= 0.375 × 500 /12.0

V1= 15.625 mL

Therefore, 15.625 mL of stock solution is required.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol