Question

What is the molarity of a solution that contains 0.978 grams of H3PO4 in 60.0 mL of solution? What is the new molarity of the H3PO4 solution if the original solution is diluted by a factor of 3? How many mL of the diluted solution could be completely neutralized by 12.5 mL of 0.545 M NaOH?

Answer #1

**Molarity of solution(M1) = (w/M)*(1000/V in
ml)**

**
= (0.978/98)*(1000/60)**

** =
0.166 M**

**volume of solution (V1) = 60 ml**

**volume of diluted solution(V2) = 60*3 = 180
ml**

**concentration of diluted solution(M2)
=(0.978/98)*(1000/180)**

**
= 0.0554 M**

**no of mol of H3PO4 present in solution = 0.978/98 = 0.01
mol**

**no of mol of NaoH added = 12.5*0.545 = 6.8125
mmol**

**1 mol H3PO4 = 3 mol NaOH**

**no of mol of h3po4 reacted = (6.8125/3*10^-3) = 0.00227
mol**

**volume of dilute H3PO4 reacted =
0.00227*10^3/0.0554**

**
= 41 ml**

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