Calculate the hydronium ion, H3O+ , and hydroxide ion, OH− , concentrations for a 0.0440 M HCl solution. [H3O+]= M [OH−]=

[H3O+] = 0.0440M

[OH-] = 2.27x10⁻¹³M

Explanation:

Based on the equilibrium of water:

2H2O ⇄ H3O+ + OH-

K = 1x10⁻¹⁴ = [H3O+] [OH-] (1)

We can determine [OH-] with [H3O+] or vice versa

As HCl is a strong acid, [HCl] = [H3O+]

That means:

[H3O+] = 0.0440M

And [OH-] is:

1x10⁻¹⁴ / [H3O+] = [OH-]

2.27x10⁻¹³M = [OH-]

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

i think the answer is terrae but not 100 percent sure. ▓▬▬ §