Cal11 calculator
Search calculator...

Calculate Oh and Ph for 0.035 M Na2s Solution

Reviewed by Calculator Editorial Team

This calculator determines the hydroxide ion concentration (OH⁻) and pH for a 0.035 molar sodium sulfide (Na₂S) solution. Sodium sulfide is a strong electrolyte that dissociates completely in water, providing hydroxide ions.

Introduction

When sodium sulfide (Na₂S) dissolves in water, it dissociates completely into sodium ions (Na⁺) and sulfide ions (S²⁻). The sulfide ions then react with water to form hydroxide ions (OH⁻) and hydrogen sulfide (H₂S):

Na₂S + H₂O → 2Na⁺ + S²⁻ + 2H₂O
S²⁻ + H₂O → HS⁻ + OH⁻
HS⁻ + H₂O → H₂S + OH⁻

This reaction produces two hydroxide ions for each sulfide ion, resulting in a higher hydroxide concentration than would be expected from a 1:1 dissociation.

Calculation Method

The hydroxide concentration can be calculated using the following steps:

  1. Determine the initial concentration of sulfide ions (S²⁻) from the Na₂S concentration.
  2. Calculate the hydroxide concentration from the sulfide ion reaction.
  3. Convert the hydroxide concentration to pH using the pOH and pH relationship.
[OH⁻] = 2 × [Na₂S] (since each Na₂S produces 2 OH⁻ ions)
pOH = -log[OH⁻]
pH = 14 - pOH

For a 0.035 M Na₂S solution:

[OH⁻] = 2 × 0.035 M = 0.07 M
pOH = -log(0.07) ≈ 1.154
pH = 14 - 1.154 ≈ 12.846

Example Calculation

Let's calculate the OH⁻ concentration and pH for a 0.035 M Na₂S solution:

  1. Initial Na₂S concentration: 0.035 M
  2. OH⁻ concentration: 2 × 0.035 M = 0.07 M
  3. pOH: -log(0.07) ≈ 1.154
  4. pH: 14 - 1.154 ≈ 12.846

The solution has a pH of approximately 12.85, indicating it is strongly alkaline.

FAQ

Why does Na₂S produce more OH⁻ than expected?
Na₂S dissociates completely in water, and the resulting sulfide ions react with water to produce two hydroxide ions per sulfide ion, leading to a higher hydroxide concentration than a 1:1 electrolyte.
Is the pH calculation accurate for all Na₂S concentrations?
Yes, the calculation is accurate for all concentrations of Na₂S, assuming complete dissociation and the reaction with water proceeds to completion.
What factors could affect the actual pH of a Na₂S solution?
Temperature, the presence of other ions, and the purity of the solution could potentially affect the actual pH, though these factors are typically negligible for basic calculations.