n/takeAssignment/takeCovalentActivity.do?locator assignment-take&takeAssignment;

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n/takeAssignment/takeCovalentActivity.do?locator assignment-take&takeAssignment; SessionLocator-assignment-tak Suggested Sites D Notifications-sw. G Google D Web Slice Gallery , imported From Safari Logn to MyAOP eon e, uary-BL s SAP The following question refers to the following system: A 1.0-liter solution contains 0.25 M HF and 1.30 M NaF (K, for HF is 7.2 x 10-1 If one adds 0.30 liters of 0.020 M KOH to the solution, what will be the change in pH? 0.01 0.19 0.73 3.87 Submit Answer Try Another Version 2 item attempts remaining 3 4 5

Explanation / Answer

mol of KOH added = 0.02M *0.3 L = 0.006 mol

HF will react with OH- to form F-

Before Reaction:

mol of F- = 1.3 M *1.0 L

mol of F- = 1.3 mol

mol of HF = 0.25 M *1.0 L

mol of HF = 0.25 mol

after reaction,

mol of F- = mol present initially + mol added

mol of F- = (1.3 + 0.006) mol

mol of F- = 1.306 mol

mol of HF = mol present initially - mol added

mol of HF = (0.25 - 0.006) mol

mol of HF = 0.244 mol

Ka = 7.2*10^-4

pKa = - log (Ka)

= - log(7.2*10^-4)

= 3.143

since volume is both in numerator and denominator, we can use mol instead of concentration

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 3.143+ log {1.306/0.244}

= 3.871

Answer: 3.87

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