Calculate the pH of a solution which contains 30 mL of acetic acid and 10 mL of 0.1M NaOH. Twenty mL of salt was formed and 10 mL of un- neutralized acid remains. The ratio of salt to acid is 10:20. The Ka for acetic acid is 1.26 x 10^-5.
Answer (1)
4.491 Calculate the moles of acetic acid (CH₃COOH) in the solution.Given that the concentration of acetic acid is not provided, we assume it is in excess and focus on the moles of NaOH. However, we need to know the initial volume of acetic acid to find its concentration. Let's denote the concentration of acetic acid as C (in M). The moles of acetic acid in 30 mL is:Moles of CH₃COOH=C×0.030 LMoles of CH₃COOH=C×0.030L2 Calculate the moles of NaOH in the solution.Using the concentration of NaOH:Moles of NaOH=0.1 M×0.010 L=0.001 molMoles of NaOH=0.1M×0.010L=0.001mol3 Determine the moles of acetic acid that remain un-neutralized.Since 10 mL of un-neutralized acetic acid remains, the moles of un-neutralized acetic acid is:Moles of un-neutralized CH₃COOH=C×0.010 LMoles of un-neutralized CH₃COOH=C×0.010L4 Calculate the moles of salt (sodium acetate, CH₃COONa) formed.From the problem, the ratio of salt to acid is 10:20, which means for every 10 moles of salt, there are 20 moles of un-neutralized acid. Therefore, if we denote the moles of salt as xx, we have:xC×0.010=1020⇒x=C×0.010C×0.010x = 2010 ⇒x=C×0.0105 Set up the equilibrium expression using the Henderson-Hasselbalch equation.The pH can be calculated using the formula:pH=pKa+log([salt][acid])pH=pKa+log( [acid][salt] )Where:pKa=−log(1.26×10−5)≈4.79pKa=−log(1.26×10 −5 )≈4.79The concentration of salt isx0.020=C×0.0100.020=C20.020x = 0.020C×0.010 = 2C and the concentration of un-neutralized acid is C×0.0100.010=C0.010C×0.010 =C.6 Substitute the values into the Henderson-Hasselbalch equation.pH=4.79+log(C2C)=4.79+log(12)pH=4.79+log( C2C )=4.79+log( 21 )pH=4.79−0.301=4.49pH=4.79−0.301=4.49
