Asolution is made by mixing exactly 500 ml of 0.156 m naoh with exactly 500 ml of 0.100 m ch3cooh. calculate the equilibrium concentration of the species below. ka of ch3cooh is 1.8 × 10−5

[h+]

× 10 m
enter your answer in scientific notation.

[oh−]

m

[ch3cooh]

× 10 m
enter your answer in scientific notation.

[na+]

m

[ch3coo−]

m

The hydrogen ion concentration is .

Explanation:

From the given,

Volume of NaOH = 500 ml = 0.500L

Moarity of NaOH = 0.156M

The number of moles of NaOH is calculated is a follows.

Substitute the given values.

Volume of = 500 ml = 0.500L

Moarity of = 0.100M

The number of moles of is calculated is a follows.

Substitute the given values.

NaOH is a strong electrolyte

The number of moles NaOH left in the solution is

Therefore, hydroxide ion concentration is 0.028 mol

The hydrogen concentration

Therefore, hydrogen ion concentration is ,

i'm assuming that it's c.) red bank..

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