Concentrated HCl is 98% and has density 1.46g/cm 3 what volume of
concentrated acid required to make 2L of 0.25M HCl .

12.64 cm^3 to the nearest hundredth

Explanation:

1 L of 0.25M of HCl contains 0.25 * (1.008 * 35.45) g of HCl

= 9.1145 g HCl.

So 2 litres contain 18.229 g HCl.

So we need the volume of concentrated acid which contains 18.229 g HCl.

1.46 * 0.98 = 1.4308 g are in  1 cm^3

So 18.229 g are in  18.229/1.4308

= 12.6404 cm^3 (answer).

0.4058L of 1.0M H3PO4

0.2192L of 1.5M NaOH

Explanation:

The pKa of the H3PO4 / H2PO4- buffer is 2.12

To solve this question we must use H-H equation for this system:

pH = pKa + log [H2PO4-] / [H3PO4]

2.75 = 2.12 +  log [H2PO4-] / [H3PO4]

0.63 = log [H2PO4-] / [H3PO4]

4.2658 = [H2PO4-] / [H3PO4] (1)

Where [] could be taken as the moles of each reactant

As you have H3PO4 solution, the reaction with NaOH is:

H3PO4 + NaOH → H2PO4- + Na+ + H2O

As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution

The moles of H3PO4 are:

[H3PO4] = Moles H3PO4 - Moles NaOH

And for H2PO4-:

[H2PO4-] = Moles NaOH added

Replacing in (1):

4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]

4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH

4.2658 Moles H3PO4 = 5.2658 moles NaOH (1)

In volume:

0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M

0.625 = Mol H3PO4 + 0.6667 Moles NaOH (2)

Replacing (2) in (1):

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667

4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)

4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4

12.1641 mol H3PO4 = 4.9367

Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =

0.4058L of 1.0M H3PO4

And:

0.625L - 0.4058L =

0.2192L of 1.5M NaOH