Suppose you prepare a buffer by adding 0.157 mol of NaH2PO4 and 0.966 mol of Na2HPO4 to 1.0 L of water. What is the pH of the solution?

pH = 7,0

Explanation:

The buffer producing by the mixture of H₂PO₄⁻ and HPO₄²⁻ is:

H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺; pka: 7,21

Using the Henderson-Hasselbalch formula:

pH = pka + log₁₀ A⁻ / HA

Where: A⁻ is the conjugate base (HPO₄²⁻) and HA is the weak acid (H₂PO₄⁻)

Replacing the formula:

pH = 7,21 + log₁₀ HPO₄²⁻ / H₂PO₄⁻

As the moles of HPO₄²⁻ are 0,0966 and moles of H₂PO₄⁻ are 0,157:

pH = 7,21 + log₁₀  0,0966 / 0,157

pH = 7,0

I hope it helps!

beta decay is the answer

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.