Question

QUESTION 17 A solution is prepared by dissolving 0.23 mol of hydrofluoric acid ($K_a = 6.8 \times 10^{-4}$) and 0.27 mol of sodium fluoride in water to yield 1.00 L of solution. What is the pH after the addition of 0.05 mol of HCl (Assume that the volume does not change)? $\bigcirc$A 3.23 $\bigcirc$B 3.11 $\bigcirc$C 3.27 $\bigcirc$D 3.06 $\bigcirc$E 4.01

          QUESTION 17
A solution is prepared by dissolving 0.23 mol of hydrofluoric acid ($K_a = 6.8 \times 10^{-4}$) and 0.27 mol of sodium fluoride in water to yield 1.00 L of solution. What is the pH after the addition of 0.05 mol of HCl (Assume that the volume does not change)?
$\bigcirc$A 3.23
$\bigcirc$B 3.11
$\bigcirc$C 3.27
$\bigcirc$D 3.06
$\bigcirc$E 4.01

QUESTION 17
A solution is prepared by dissolving 0.23 mol of hydrofluoric acid (Ka = 6.8 × 10^-4) and 0.27 mol of sodium fluoride in water to yield 1.00 L of solution. What is the pH after the addition of 0.05 mol of HCl (Assume that the volume does not change)?
◯A 3.23
◯B 3.11
◯C 3.27
◯D 3.06
◯E 4.01

Added by William M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Muhammad Ahsan

Step 1

To do this, we need to divide the number of moles of each compound by the total volume of the solution: [H+] = 0.23 mol / 1.00 L = 0.23 M [F-] = 0.27 mol / 1.00 L = 0.27 M Show more…

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A solution is prepared by dissolving 0.23 mol of hydrofluoric acid (Ka = 6.8x10^-4) and 0.27 mol of sodium fluoride in water to yield 1.00 L of solution. What is the pH after the addition of 0.05 mol of HCl (Assume that the volume does not change)? A) 3.23 B) 3.11 C) 3.27 D) 3.06