6. (5 points) NaCT is added slowly to a solution that is 0.010 M each in aqueous

Question

6. (5 points) NaCT is added slowly to a solution that is 0.010 M each in aqueous Cu'.Ag.and Au The Kip's for Cuci.AgC,and AuCl are 19x 10.18 x 01and x T, respectively. larger . kspe more sila bU Which compound will preciatate first? x10-13 which IS Avci,"wide ma ke sense since melal the heavies 7. (15 points) A solution of 20,.00 ml of Q,.1175 M HCI (aq) is titrated with 0.1650 M NaOH (aq). Hcl is a strong acid whchnensit Copletely dsscet A. Calculate the initial pH of the HCI solution. kd means tuat l Hescorupletely..eutralne d I mole o B. what volume of NaOH (aq) will be required to reach the equivalence point? = o. 61424 Shbw bu Show nuel.co235ous HCI Ne 6 lloSD C. What is the pH of the solution at the equivalence poinp AT Huu t qui Valince pomt both acid CHed and bes Na0H are con sumed completely goluhon belous neutral when itls m nutral point 4

Explanation / Answer

Q6.

you are correct, you can compare Ksp directly sinece salts are XCl type

then large Ksp imply more soluble

Q7

a)

pH of HCl is corret

b)

Macid*Vacid = Vbase*Mbase

Vbase= Macid*Vacid /Vbase = 0.1175*20/0.165 = 14.24 mL required

c.

in equivalence point p H= 7 , true, you are correct

d.

mmol of NaOH = MV = 0.165*6.5 = 1.0725

mmol of HCl= MV = 0.1175*20 = 2.35

mmol of H+ = 2.35 - 1.0725 = 1.28

Vtotal = 6.5+20 = 26.5

[H+] = 1.28/26.5 = 0.0483

p´H = -log(0.0483) = 1.32

E

mmol of NaOH = MV = 0.165*27.55= 4.54575

mmol of HCl= MV = 0.1175*20 = 2.35

mmol of OH- = 4.54575- 2.35 = 2.20

Vtotal = 27.55+20 = 47.55

[OH-] = 2.20/47.55 = 0.0462

pOH = -log(0.0462) = 1.34

pH = 14-1.34 = 12.66

Verifity total volume

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