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2 years ago

If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral

Chemistry

1 answer:

raketka [301]2 years ago

4 0

Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

If no. of millimoles of acid > that of base; the solution is acidic.

If no. of millimoles of acid = that of base; the solution is neutral.

If no. of millimoles of acid < that of base; the solution is basic.

We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).

So, the solution is: basic.

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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo

garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c) 0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

1 mole 3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

10 ml 17.50 ml

(x) M 0.200 M

Molarity = $\frac{0.2*17.5}{1000}$

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= $0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}$

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration = $\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }$

Molar Concentration = $\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000$

Molar Concentration = 0.166 M

∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$