Answer:
- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL
Explanation:
Hello,
In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:
- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.
Regards.
Explanation:
Volume is the quantity of three-dimensional
Answer:
97.9 g/mol
Explanation:
<em>There is some info missing. I think this is the original question.</em>
<em>An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0600 M NaOH solutions. When the titration reaches the equivalence point, the chemist finds he has added 95.9 mL of NaOH solution.
</em>
<em>
Calculate the molar mass of the unknown acid.</em>
Let's consider the neutralization between a generic triprotic acid and NaOH.
H₃X + 3 NaOH → Na₃X + 3 H₂O
The moles of NaOH that reacted are:
0.0600 mol/L × 0.0959 L = 0.00575 mol
The molar ratio of NaOH to H₃X is 3:1. The moles of H₃X are 1/3 × 0.00575 mol = 0.00192 mol.
The molar mass of the acid is:
0.188 g / 0.00192 mol = 97.9 g/mol
