Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to (a) 42.0 mL of 0.320 M NaOH(aq). (b) 22.0 mL of 0.420 M NaOH(aq).

a) pH = 11.5

b) pH= 3

Explanation:

a) lets calculate the number of moles of each reactant

moles of HCl = 32/1000 * 0.32 = 0.01024 mole

moles of NaOH = 42/1000 * 0.32 = 0.01344 moles

1 moles of HCl reacts with 1 moles of NaOH , so 0.01024 mole of HCl should react with 0.01024 moles of NaOH , but there is some excess NaOH.

excess NaOH= 0.01344 -0.01024 = 0.0032 moles

[H+]=

pH= -log [3.125*10^-12) = 11.5

b)  moles of NaOH = 0.00924

excess HCl present = 0.01024 - 0.00924 =.001

so excess [H+] = 0.001

pH= -log( 0.001) = 3

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3