A is a solution of trioxonitrate(v) acid, HNO3, of unknown concentration. B is standard solution of sodium hydroxide containing 4.00g per dm3 of solution. 25cm3 portions solution B required an average of 24.00cm3 of solution A for neutralization, using 2drops of methyl orange.

a: write a balanced equation for the reaction?
b:Calulate the concentration of solution B in moles per dm3 ?
c: calculate the concentration of solution A in moles per dm3
d: calculate the concentration in g/dm3 of HNO3 in solution A. (H=1, N=14, O=16, Na=23)

I would be much happier if you just called HNO3 as nitric acid and Pb(NO3)2 as lead(II) nitrate.

a.
HNO3 + NaOH ==> NaNO3 + H2O

b.
mols NaOH = 4/molar mass NaOH = ?

c.
M NaOH(soln B) = mols/cm3 = ?
cc(HNO3) x M(HNO3) = cc(NaOH)x M(NaOH. Substitute and solve for M HNO3.

d.
I'll leave this for you.

conc. of solution b is 0.1 molperdm3conc. of solution a inmolperdm3 is 0.01 conc. in gperdm3 of hno3 in solution a is 0.63g/dm3

Can u paste how the solution was solved

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CHEMISTRY

a: The balanced equation for the reaction between trioxonitrate(v) acid (HNO3) and sodium hydroxide (NaOH) is:

HNO3 + NaOH → NaNO3 + H2O

b: To calculate the concentration of solution B (NaOH) in moles per dm3 (M), we need to use the formula:

Molarity (M) = moles of solute (NaOH) / volume of solution (dm3)

Given that the standard solution of sodium hydroxide contains 4.00g per dm3 of solution, we need to convert the mass of sodium hydroxide to moles:

Moles of NaOH = mass of NaOH (g) / molar mass of NaOH

The molar mass of NaOH is 23 (atomic mass of Na) + 16 (atomic mass of O) + 1 (atomic mass of H) = 40 g/mol.

Moles of NaOH = 4.00 g / 40 g/mol = 0.1 mol

Since the volume of solution B used for neutralization is 25 cm3, we need to convert it to dm3:

Volume of solution B (dm3) = 25 cm3 / 1000 = 0.025 dm3

Now we can calculate the concentration (M) using the formula:

M = 0.1 mol / 0.025 dm3 = 4.00 mol/dm3

Therefore, the concentration of solution B (NaOH) is 4.00 mol/dm3.

c: To calculate the concentration of solution A (HNO3) in moles per dm3 (M), we can use the balanced equation from part (a) and the volume of solution A used for neutralization. Since equal volumes of the acid and base react to neutralize each other, we can conclude that:

Volume of solution A (dm3) = Volume of solution B (dm3) = 0.025 dm3

Therefore, the concentration of solution A (HNO3) is also 4.00 mol/dm3.

d: To calculate the concentration of HNO3 in g/dm3 in solution A, we need to consider the molar mass of HNO3 and use the concentration in moles per dm3 from part (c).

The molar mass of HNO3 is 1 (atomic mass of H) + 14 (atomic mass of N) + 3x16 (3 times the atomic mass of O) = 63 g/mol.

To convert the concentration (in moles per dm3) to grams per dm3, we can use the formula:

Concentration (g/dm3) = Concentration (mol/dm3) x molar mass (g/mol)

Concentration (g/dm3) = 4.00 mol/dm3 x 63 g/mol = 252 g/dm3

Therefore, the concentration of HNO3 in solution A is 252 g/dm3.

Correct

A is a solution of trioxonitrate (iv) acid (nitric acid) of unknown concentration. B is a standard solution containing 4.00g, 25cm3 portion of solution B requires an average of 24.00cm3 of solution A for complete neutralization

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