The pOH of a solution is 13.28. What is the [OH]

1 answer:

3 0

Say that this is the concentration of OH- ions in solution. All you have to do is to get the negative log of the [OH-] to get the pOH. After getting the pOH, get the pH of the solution.

pOH = -log[OH-]
= -log(1.0 * 10^-5)
= 5.00

This is actually the pOH value, not the pH. To get the pH, just subtract the pOH from 14.

pH = 14 - pOH
= 14.00 - 5.00
= 9.00

The value 2.5 x 10^-11 M is the concentration of OH- ions in solution. To get the pH, get the pOH first.

pOH = -log[OH-]
= -log(2.5 x 10^-11)
= 10.6

Then, subtract this from 14 to get the pH. The pH is 3.40.

Note: When you determine the pOH or pH of the sample, always make sure you enter -log in the calculator so you'll get a positive result.

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sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$