Question

how many ml of 3.000 m naoh solution are needed to react with 2.000 g h₃po₄?
the molar mass of h₃po₄ is 97.98 g.
h₃po₄ + 3 naoh → 3 h₂o + na₃po₄
this is a titration problem, and we are given enough information to calculate the
moles of the acid.
1 mol
(2.000 g)(----------) = 1 mol h₃po₄
97.98 g
2 mol naoh
(3 mol h₃po₄)(-------------------) = 4 mol naoh
1 mol h₃po₄
1 l
(5 mol naoh)(----------------) = 6 l naoh
7 mol naoh
8 ml
(9 l naoh)(----------) = 10 ml naoh
1 l
a. 3.000 b. 2.500 c. 10.00 d. 1.000 e. 2.000
f. 9.000 g. 6.000 h. 8.000 i. 0.02500
j. 47.00 k. 5.00 l. 50.00 m. 25.0 n. 0.02041
o. 97.98 p. 30.62 q. 6.572 r. 0.06124 s. 262.87

Explanation:

Step1: Calculate moles of \(H_3PO_4\)

$$n_{H_3PO_4}=\frac{m}{M}=\frac{2.000\ g}{97.98\ g/mol}\approx0.02041\ mol$$
So, blank 1 is \(0.02041\) (N).

Step2: Determine mole ratio from reaction

From \(H_3PO_4 + 3NaOH
ightarrow3H_2O+Na_3PO_4\), the mole ratio of \(NaOH\) to \(H_3PO_4\) is \(3:1\). So blank 2 is \(3\) (A).

Step3: Use moles of \(H_3PO_4\)

We use the moles of \(H_3PO_4\) calculated in step 1. So blank 3 is \(0.02041\) (N).

Step4: Calculate moles of \(NaOH\)

$$n_{NaOH}=n_{H_3PO_4}\times3=0.02041\ mol\times3 = 0.06124\ mol$$
So blank 4 is \(0.06124\) (R).

Step5: Use moles of \(NaOH\)

We use the moles of \(NaOH\) calculated in step 4. So blank 5 is \(0.06124\) (R).

Step6: Calculate volume of \(NaOH\) solution (in L)

Using \(V=\frac{n}{C}\), with \(C = 3.000\ M\) (so blank 7 is \(3.000\) (A))
$$V_{NaOH}=\frac{0.06124\ mol}{3.000\ mol/L}\approx0.02041\ L$$
So blank 6 is \(0.02041\) (I).

Step7: Convert volume to mL

$$V_{NaOH}(mL)=0.02041\ L\times1000\ mL/L = 20.41\ mL\approx20.41\ mL$$
But if we follow the blanks:
For blank 9, we use the volume in L (\(0.02041\) (I)) and for blank 8 (\(1000\)) and blank 10 (\(20.41\) but looking at options, if we assume some rounding in steps)
If we recalculate:
\(n_{H_3PO_4}=\frac{2.000}{97.98}\approx0.02041\)
\(n_{NaOH}=0.02041\times3 = 0.06123\)
\(V_{NaOH}(L)=\frac{0.06123}{3.000}=0.02041\)
\(V_{NaOH}(mL)=0.02041\times1000 = 20.41\) but if we consider the options and proper significant - figures (since \(2.000\ g\) has 4 sig - figs, \(3.000\ M\) has 4 sig - figs)
Another way:
\(n_{H_3PO_4}=\frac{2.000}{97.98}\)
From reaction \(n_{NaOH}=3\times n_{H_3PO_4}\)
\(V_{NaOH}(L)=\frac{3\times2.000}{97.98\times3.000}\)
\(V_{NaOH}(mL)=\frac{3\times2.000\times1000}{97.98\times3.000}\approx20.41\) but if we check the options again, maybe there was a miscalculation in the problem - setup (if we assume \(n_{H_3PO_4}=\frac{2.000}{97.98}\approx0.02041\), \(n_{NaOH}= 0.06123\), \(V=\frac{0.06123}{3}=0.02041\ L = 20.41\ mL\) but if we use the blanks:

1. \(n_{H_3PO_4}=\frac{2.000}{97.98}\approx0.02041\) (N)
2. Mole ratio \(3\) (A)
3. \(n_{H_3PO_4}=0.02041\) (N)
4. \(n_{NaOH}=0.06124\) (R)
5. \(n_{NaOH}=0.06124\) (R)
6. \(V_{NaOH}(L)=\frac{0.06124}{3.000}=0.02041\) (I)
7. \(C = 3.000\) (A)
8. \(1000\)
9. \(0.02041\) (I)
10. \(20.41\approx20.41\) but if we assume in the problem - setup (maybe a typo in molar mass or concentration)

If we do it as:
\(n_{H_3PO_4}=\frac{2.000}{98}\approx0.02041\) (using \(M = 98\) approx)
\(n_{NaOH}=0.02041\times3=0.06123\)
\(V_{NaOH}(L)=\frac{0.06123}{3}=0.02041\)
\(V_{NaOH}(mL)=20.41\)

Answer:

1. N. \(0.02041\)
2. A. \(3.000\)
3. N. \(0.02041\)
4. R. \(0.06124\)
5. R. \(0.06124\)
6. I. \(0.02041\)
7. A. \(3.000\)
8. \(1000\) (not in options, but unit - conversion factor)
9. I. \(0.02041\)
10. \(20.41\) (not in options, but if we assume calculation with \(M = 98\) and \(C = 3\) exactly: \(\frac{2\times3\times1000}{98\times3}\approx20.41\))