I need help calculating the ph after addition of strong base naoh 1 M of 0.100 m

Question

I need help calculating the ph after addition of strong base naoh 1 M of 0.100 mL to my buffer. My acid concentration is 0.001226 M My base concentration is 0.000774 M
I need to use Henderson hesselbach equation to compute the pH I need help calculating the ph after addition of strong base naoh 1 M of 0.100 mL to my buffer. My acid concentration is 0.001226 M My base concentration is 0.000774 M
I need to use Henderson hesselbach equation to compute the pH My acid concentration is 0.001226 M My base concentration is 0.000774 M
I need to use Henderson hesselbach equation to compute the pH

Explanation / Answer

NOTE, you did NOT stated amount of volume of buffer so I will assume 1 Liter of buffer solution, (simply change L for your volume)

Find moles of weak acid:

mol = MV = 0.001226 *1 = 0.001226 mol of acid

Find moles of conjugate base

mol = MV = 0.000774 *1 = 0.000774 mol of base

After adding:

mol of NaOH = MV = (1)(0.1*10^-3) = 0.0001

so..

after addition:

Find moles of weak acid:

initial - base reacted = 0.001226 -0.0001 = 0.001126 mol of weak acid left:

Find moles of conjugate base

initial + base added = 0.000774 + 0.0001 = 0.000874 mol of weab base formed

so...

pH = pKa + log(conjugate base / weak acid)

note that we need pKa for this... no pKa is stated so simply add your pKa value

pH = pKa + log(0.000874 / 0.001126 )

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