Consider a solution prepared by mixing the following: 50.0 ml of 0.100 m na3po4100.0 ml of 0.0500 m koh200.0 ml of 0.0750 m hcl50.0 ml of 0.150 m nacndetermine the volume of 0.100 m hno3 that must be added to this mixture to achieve a nal ph value of 7.21.

you must add 50 mL

Explanation:

Hi

KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.

NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.

HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.

The acid reacts with the bases leaving 2.5 mmol unreacted.

Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.

The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12

5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21

Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.

To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.

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sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.