(a) We can use the following relationship for dilution – V1S1 = V2S2, V1=10.0 ml

Question

(a)

We can use the following relationship for dilution –

V1S1 = V2S2, V1=10.0 ml of KOH and S1= 3.00 M and V2 = 250 ml of KOH, S2=?

S2= V1S1/V2 = 10*3/250 = 0.12 M

Now, for the acid-base reaction, we can use the following relationship -

V3S3 = V4S4, where, V3=38.5 ml, the volume of KOH; S3= 0.12 M, molarity of the diluted KOH solution.

V4=10 ml, the volume of H3PO4 and S4 = ?, molarity of H3PO4.

S4 =V3S3/V4 = 38.5*0.12/10 M = 0.462 M is the molarity of H3PO4.

(b)

By definition, 1 M, 1L contains =1 mol

0.462 M, 10 ml contains H3PO4 = 0.462*10/1000 mol

= (97.994 g/mol )*0.462*10/1000(mol) = 0.4527 g of H3PO4 was in the initial sample.

Explanation / Answer

3. Suppose that 10.0 mL. of 3.00 M KOH is transferred to a 250.0-mL volumetric flask and diluted to 250.0 mL. It was found that 38.5 mL of this diluted solution was needed to react with 10.0 mL of a phosphoric acid (HsPO) solution according to the reaction: KJ04(aq) 3H20(4 + 3KOH(ag) H104(ag) + a) Calculate the molarity of the phosphoric acid solution. b) What mass of phosphoric acid is in the initial sample of H PO? Total 40

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