First find the no. of moles of NaOH :
<span>30/1000 = 0.3 dm3 so no. of moles = 0.3*0.5 = 0.15 moles </span>
<span>as NaOH reacts with HNO3 in a ratio of one to one, there must have been 0.15 moles of HNO3 too </span>
<span>moles/volume = concentration </span>
<span>volume= 15/1000 = 0.15 dm3 </span>
<span>concentration = 1.15/0.15 = 1 mol.dm-3 </span>
<span>The quicker way would be to realize that you used twice as much NaOH so the HNO3 had to be twice as strong</span>
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
Answer:
s an example, the ground state configuration of the sodium atom is 1s22s22p63s1, as deduced from the Aufbau principle (see below). The first excited state is obtained by promoting a 3s electron to the 3p orbital, to obtain the 1s22s22p63p1 configuration, abbreviated as the 3p level.
Explanation:
Find the [OH-] in the solution. The pH is 9.5, so the pOH is 14 - 9.5 = 4.5.
[OH-] = 10^-4.5 M
Now use the dilution equation to find the new [OH-] after the volume is reduced from 150 mL to 50 mL:
M1V1 = M2V2
M1 = 10^-4.5 M
V1 = 150 mL
M2 = ?
V2 = 50 mL
(10^-4.5)(150) = M2(50)
M2 = 9.5 x 10^-5 M ≈ 1 • 10^-4 (We can only use one sig fig, because the pH was given to one decimal place.)
Now use this [OH-] to find pOH:
pOH = -log(1 x 10^-4) = 4.0
14 - pOH = pH, so the expected pH for the new solution is 10.
