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NaOH(s) was added to 1.0 L of HNO3(aq) 0.48 M. 

Calculate [H3O+] in the solution after the addition of 0.11 mol of NaOH(s).
you need to consider the volume of the solution is constant, which means that we are not combining two solutions together like in the other neutralization problems. 

2,,Calculate [H3O+] in the solution after the addition of 0.89 mol of NaOH(s)

 you need to need to find the concentration of the [OH-] then use the Kw equation to find the hydronium concentration.

Apr 4th, 2015

1. since the volume of sodium hydroxide is not specified, we will have to assume that the volume remains at a constant 1L after addition of NaOH. In this case, knowing that NaOH is a strong base and HNO3 is a strong acid, they will neutralize each other to form water. 

H3O+(aq) + OH-(aq) --> 2H20(l) 

0.48 moles HNO3 - .11 moles of NaOH = 0.37 moles H+ left over 
Usually, you would be given the volume of NaOH added, since the [H3O+] is dependent on volume, such that: 

.37 mol / (1L + V of NaOH added) = [HNO3+] 

Since we are assuming the volume is constant at 1L, the [HNO3+] is equal to 0.37M after addition of 0.11 moles of NaOH. 

2. The same concept applies; except that this time, the 0.48 moles of HNO3+ is completely neutralized by the moles of NaOH added, since the moles of NaOH are greater than 0.48. 

Therefore, we are left with: 

0.89 mol NaOH - 0.48 moles HNO3+ = .41 moles OH- 

Once again assuming that the volume is held constant at 1L, the concentration of OH- is 0.41M. Since we need to calculate [H3O+] and this is a strong acid/strong base titration, we know that: 

The water dissociation constant, Kw, is equal to 1.0x10^-14 
Kw = [H3O+][OH-] 
Since we have [OH-], we divide both sides by [OH-] to get [H3O+] by itself. 

[H3O+] = 1.0x10^-14/0.41 = 2.4x10^-14M 

Apr 4th, 2015

Apr 4th, 2015
Apr 4th, 2015
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