Some problems... (1 Viewer)

[Aerlinn]

Some quick questions...
25.00mL of a 0.100M solution of HCl is added to 25.00mL of a 0.180M solution of NaOH. The concentration of OH-(aq) remaining in the solution, in M, is?
50.00mL of a 0.020M solution of Ba(OH)2 is added to 50.00mL of a 0.060M solution of HNO3. THe hydrogen ion concentration in the resultant solution, in mole per litre, is?

Some insight would be great

 

[Ea22.007]

one way to solve this is just work out the moles of H+ and OH- in each soln.
to do this u have to multiply the concentrations by the amount of solution in litres so for the first soln its 0.025L*0.1mol/L

then as they will neutralize each other, and there is an excess of OH- u can just go
(mols of OH) minus the (mols of H) to get the excess of OH left in the soln.

then to get its concentration u just divide that by 0.05liters to the the concentration of OH ions in mols per litre

 

[Ea22.007]

its a similar process for the second question

 

[xiao1985]

ditto for first one...

2nd one is slightly different... for each mole of ba oh2, you have 2 moles of oh-

so you will have:

no of moles of oh-:0.05x0.02x2 =0.2 moles

 

[elseany]

can someone post up the answers?

mine were

a) 0.04M
b) 0.01M