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In: Chemistry

A 25.0 mL sample of a 0.150 M weak acid, Ka = 4.6 x 10−5 ,...

A 25.0 mL sample of a 0.150 M weak acid, Ka = 4.6 x 10−5 , is titrated with 0.200 M NaOH. Calculate the pH when 10.0 mL of the base has been titrated into the acid.

Expert Solution

no of moles of acid HA = molarity * volume in L

= 0.15*0.025 = 0.00375 moles

no of moles of NaOH = molarity * volume in L

= 0.2*0.01 = 0.002mole

HA + NaOH ----------------> NaA + 2H2O

I 0.00375 0.002 0

C -0.002 -0.002 0.002

E 0.00175 0 0.002

Ka = 4.6*10^-5

PKa = -logKa

= -log4.6*10^-5

= 4.3372

PH = Pka + log[NaA]/[HA]

= 4.3372 + log0.002/0.00175

= 4.3372 + 0.05799

= 4.3951 >>>>>answer

queen_honey_blossom answered 2 years ago

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