pH of solution q (1 Viewer)

[username_x]

What is the pH of a solution prepared by mixing 30mL of 0.1M Nitric acid solution with 10mL of 0.1M barium hydroxide solution?

I keep getting this wrong.. can someone please help

 

[silence--]

number of moles of HNO3 = conc*volume = 0.03*0.1 = 0.003
similary number of moles Ba(OH)2 = 0.001

now 2HNO3 + Ba(OH)2 ---> Ba(NO3)2 + 2H2O

mole ratio of barium hydroxide to nitric acid is 1:2
so for 0.001 moles of barium hydroxide, 0.002 moles of nitric acid will react.
so there are 0.003 - 0.002 = 0.001 leftover moles of HNO3.

concentration of leftover = 0.001 / 0.04 (new volume by adding 10mL and 30mL)
=0.025

pH = -log[H+} = 1.602

is that right?

 

[username_x]

That's the answer I got, yet the answers say it is 1.3... very confused :| I think I shall just ignore it

but just wondering, with the Ba(OH)2, does that mean that the conc for that is doubled as there are two OH- ions?

 

[b3kh1t]

What is the pH of a solution prepared by mixing 30mL of 0.1M Nitric acid solution with 10mL of 0.1M barium hydroxide solution?

I keep getting this wrong.. can someone please help

first we find the moles of H+ and OH- released.

n(HNO3) = 0.030 x 0.1 = 3x10^-3 moles
since HNO3 ionises to produce one H+ ions per molecule, then 3x10^-3 moles of H+ are present

n(Ba(OH)2) = 0.010x0.1 = 1x10^-3 moles
since Ba(OH)2 ionises to produce two OH- ions per molcule, then 2x1x10^-3 = 2x10^-3 moles of OH- are present

therefore 2x10^-3 moles of OH- will neutralise 2x10^-3 moles of H+, and 1x10^-3 moles of H+ ions will remain in the solution

now to find the pH you MUST find the concentration of the H+ using C= n/V = 1x10^-3/(0.030+0.010) = 0.025 mol/L
thus the pH is -log(0.025)= 1.6

 

[b3kh1t]

number of moles of HNO3 = conc*volume = 0.03*0.1 = 0.003
similary number of moles Ba(OH)2 = 0.001

now 2HNO3 + Ba(OH)2 ---> Ba(NO3)2 + 2H2O

mole ratio of barium hydroxide to nitric acid is 1:2
so for 0.001 moles of barium hydroxide, 0.002 moles of nitric acid will react.
so there are 0.003 - 0.002 = 0.001 leftover moles of HNO3.

concentration of leftover = 0.001 / 0.04 (new volume by adding 10mL and 30mL)
=0.025

pH = -log[H+} = 1.602

is that right?

yes except be careful of the significant figures, they may be picky and take marks off

 

[username_x]

but isn't the mole ratio 2:1 in the reaction (of acid to base)?