A buffer contains 0.50 mol NH₄⁺ and 0.50 mol NH₃ diluted with water to 1.0 L.How many moles of NaOH are required to increase the pH of the buffer to 10.00? (pK_a of NH₄⁺ = 9.25) A) 0.035 mol B) 0.15 mol C) 0.35 mol D) 0.41 mol E) 2.8 mol

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To calculate the moles of NaOH required to increase the pH of the buffer to 10.00, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) We know that pKa = 9.25 and [A-]/[HA] = 1, since the buffer is at a 1:1 ratio. Therefore, we can rearrange the equation to solve for [OH-]: pOH = 14 - pH = pKb + log([OH-]/[B]) Since pKb + pKa = 14, we can calculate pKb as 4.75. Substituting in the values, we get: 9.00 = 4.75 + log([OH-]/0.5) Solving for [OH-], we get: [OH-] = 10^(9.00 - 4.75 - log(0.5)) = 0.0137 M To convert [OH-] to the amount of NaOH required, we need to multiply by the volume of the buffer: moles of NaOH = 0.0137 mol/L * 1 L = 0.0137 mol However, this is just the amount required to increase the pH to 9.00. Since we want to increase it to 10.00, we need to repeat the calculations using a new pKa value. The pKa of NH₃ is 10.0, so the new pKb value is 4.0. Using the same calculations as above, we get: 10.00 = 4.00 + log([OH-]/0.5) [OH-] = 0.327 M moles of NaOH = 0.327 mol/L * 1 L = 0.327 mol Adding the moles of NaOH required for both calculations, we get: 0.0137 mol + 0.327 mol = 0.34 mol Thus the answer is C) 0.35 mol, with some rounding to two significant figures.

A Buffer Contains 0