QUESTION 3 Consider a solution containing 0.80 M NaF and 0.80 M HF. Calculate the moles of HF and the concentration of HF after addition of 20.0 mL of 0.40 M HCl to 60.0 mL of this buffer solution. 0.088 moles, 0.99 M 0.055 moles, 0.11 M 0.048 moles, 0.60 M 0.088 moles, 0.20 M 0.056 moles, 0.93 M 0.040 moles, 0.67 M 0.056 moles, 0.70 M

0.056moles HF and 0.70M

Explanation:

When a strong acid is added to a buffer, the acid reacts with the conjugate base.

In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:

NaF + HCl → HF + NaCl

Initial moles of NaF and HF in 60.0mL of solution are:

NaF:

0.0600L × (0.80mol / L)= 0.048 moles NaF

HF:

0.0600L × (0.80mol / L)= 0.048 moles HF

Then, the added moles of HCl are:

0.0200L × (0.40mol / L) = 0.008 moles HCl.

Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:

0.056moles HF

In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:

0.056mol HF / 0.0800L = 0.70M

it is a good conductor

explanation: