Question

If a 16.5 mL sample of 1.3 M solution of each of the following acids is reacted with 0.90 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point?

16.5 mL of HCl titrated with 0.90 M NaOH

volume of NaOH ___mL

total volume___ mL

16.5 mL of H₂SO₄ titrated with 0.90 M NaOH

volume of NaOH ___mL

total volume____ mL

16.5 mL of HClO titrated with 0.90 M NaOH

volume of NaOH ____mL

total volume____ mL

Answer 1

Solution 1.3M Concentrcution aud Lis 16.5 mL Hl titrated uth .90M NaoH urecaction Nacl HC +N40H Conc of acid x veume lin 0.0165 L O.02145 mol t no. d melus of tcl 1.3M X Se, faom alove ireactio mod of tCL So, o.0214s mel Hcl mel of NUOH heedls heecs NacH C.02145 me So, 0.0145 mo NaoH uquircdl no. et mols solute volume of, sdution lin L) Molarity So Molerity NaoH og0M no. a moL aNaOlH= O.02145 mol So, velune ot NaoH. O.02145 mol 0.S0M 0.023% L 23.8mL vdume ot Totcel valume 23.3 Naold mL 16-5 mL + 23.8mL 40.3 mL

Ci 16.5 mL ot 4s04 titrcuteel uith O.90M M40H Oceaction H SOu + Maisou t 2 HO 2 NaO H no, ot mels 3 M Hsoy D.OL65L 2 0.02145 med Ercm the alove uactron:- 1mel of H,so needs So, o.02145 mol of H2sa needs (2x 0-021u51mot of Nao4. 2 mol oh NaoH 6.0429 mol NaoH 0.o429 mol oNaoH equircol Molari y no of moles of solute Uolume of solutton (inL). volume of Nao 0 O429 mol O.90 M 47.6.mL 1 Ju.6 mL volume of Total volu me NaoH 16-5 mi t 7.6mL 64.1 mL

Ci 16.5 mL ot Hclo titratrd rith 0.90 M NaoH Uuactron tclo t Naclo +H2O NaoH no et molus of Hclo O.O165L XI.3 M O02145 mol So, From alove vreactron mol of NUOlH heeds Imel of Hclo So, o.02145 mol e Hclo needs N4o14 0.02145 mol of Molarity no ot molos of solwte velume of solution Lin LS "velume of NaoH solutton 0 02145 Wmol O-90M 30.023% L 23. mL Volume o NaoH 23- mL Total volume 165 m + 23.8 mL 40.3 mL