Question

What is the pH after the addition of 10.0 mL of 0.150 MNaOH to 25.0 mL of 0.120 MHNO3?

Answer 1

HNO₃ (nitric acid) reacts with Naon (sodium hydroxide) to form sodium nitrate and Wares No caq.) + Naokcaq. ) NaNo logo) + H₂O) Ginen, Mi = concentration of Na on = 0.150M V = Volveme L NOOH 10.0 mLop*044 M = concentration of HNO₃ = 0.120M V2 = volume of HNO₃ = 25.0m. n, moles si Naona M, XV, = 0,150 Mx 10.0m 2 an, = 11.50 millimoles M2 = moles of nNO3 = M₂ X V₂ = 0.120m x 2.5.0m2 12 = 3.00 millimoles wow, I mole of Naon reals with Imole of NO₂. (By slochiometry) : 1.5 millimoles of Noon reals with 1.5 millinoley MNO3 . . Moles of nNo, left = (3.0 - 1.5) mollinoles . .. in;= 1.5 millimales) final volume of solution = volume of + volume oud of babe . Ve 10.0 mL + 25.0mL Ne 35.0mL Scanned with CamScanner

concentration of HNO3 left = 23 [HNO] : 1.50X 10-3 moles 35.0 10-3 litre [HNO JE Jo.0429 M/ Imole HNO2 condary I molek [ht] [ HNO3) = 10.0429M ] we know that, PME -log([nt]) i. pH = -log(0.0429 M) Iph = 1.37 Scanned with CamScanner
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