What is the pOH of a solution with an H3O+concentration of 1x10-6 M

Answer : The H_3O^+ concentration and the pH of the solution is, 1\times 10^{-9}M and 9 respectively.

Explanation:

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1\times 10^{-5})

pOH=5

Now we have to calculate the pH.

pH+pOH=14pH=14-pOHpH=14-5=9

Now we have to calculate the H_3O^+ concentration.

pH=-\log [H_3O^+]

9=-\log [H_3O^+]

[H_3O^+]=1\times 10^{-9}M

Therefore, the H_3O^+ concentration and the pH of the solution is, 1\times 10^{-9}M and 9 respectively.

Explanation:

c. 75.7

i think that is the right

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