what is the molarity of a HCI solution if 25.0 ml of 0.185 M Main is required to neutralize 0.0200 L of HCI?

Molarity of HCl solution = 0.25 M

Explanation:

Given data:

Volume of NaOH= V₁ = 25.0 mL (25/1000 = 0.025 L)

Molarity of NaOH solution=M₁ = 0.185 M

Volume of HCl solution = V₂ = 0.0200 L

Molarity of HCl solution =  M₂= ?

Solution:

M₁V₁   =  M₂V₂

0.185 M ×0.025 L =  M₂ × 0.0200 L

M₂  = 0.185 M ×0.025 L / 0.0200 L

M₂  = 0.005M.L /0.0200 L

M₂  =  0.25 M

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

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