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Solved Step by Step With Explanation- Calculate pH of solutions

Questions

Part B 

0.085 mol/LRbOH and 0.105 mol/L of NaCl calculate PH= ?

Answer

Solved Step by Step With Explanation- Calculate pH of solutions

NH4NO3 ⇌ NH4+ + NO3-

Consider the dissociation of HCN in water:

Kb = (NH3)(H3O+) / NH4+

1.76 x 10^(-5) = (x)(x) / 0.265

x ≈ 0.00216 M (rounded to four decimal places)

Now, we know that Kw = [H3O+][OH-], so:

OH- ≈ 4.63 x 10^(-12) M (rounded to four decimal places)

Calculate the pH using the equation:

PART B)

In this part, you're given the concentrations of RbOH and NaCl and asked to calculate the pH of the solution.

pOH = -log[OH-]

pOH = -log(0.085)

pH ≈ 12.93

So, the pH of the solution is approximately 12.93.

Since HClO4 is a strong acid and KOH is a strong base, they will react in a 1:1 molar ratio, resulting in water (H2O) and potassium perchlorate (KClO4), which is a salt that does not affect the pH.

Use the Kw expression to calculate the concentration of H+ ions:

[H+] ≈ 4.167 x 10^(-13) M

Step 3: Calculate the pH using the equation:

Finally, let's move on to Part E.

PART E)

Consider the reaction between ClO- and I-:

ClO- + I- ⇌ HClO + I-

Given that Ka for NaClO is 4 x 10^(-8), we can calculate [H+]:

[H+] = (0.117)(4 x 10^(-8)) / (0.117) = 4 x 10^(-8) M

So, the pH of the solution is approximately 7.398.