Concentration of NaOH = 0.148 molar, M
Concentration of H3PO4 = 0.172 molar, M
Concentration x Volume will give the number of moles of solute in that volume. C*V = moles
Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.
Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl
The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:
NaOH + HCl > NaCl and H2O
This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.
The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:
(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M
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The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:
H3PO4 + 3NaOH = Na3PO4 + 3H2O
Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.
The moles of NaOH that were used is:
(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH
Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.
The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:
(0.00172 moles H3PO4)/(0.010 liters H3PO4)
Concentration of H3PO4 = 0.172 molar, M