Answer:
The concentration of HBr is 0.399M to 3 sig figs
Step-by-step explanation:
The place to start is with a balanced equation:
HBr + NaOH = H2O + NaBr
This looks balanced. We can see that one mole of HBr is needed for every 1 mole of NaOH. Titration means that we will have an endpoint that neutralizes the HBr after sufficient NaOH is added. In this case we need to add the same number of moles of NaOH as we already have in the 19.32 ml of HBr. We are missing the concentration of HBr, but we have the information we need to calculate the moles of NaOH that were added to obtain the endpoint, so lets calculate the moles of NaOH that were used.
Moles of NaOH in 42.78 mL of 0.180M NaOH
M means Molar, which is defined as moles/liter. Use that unit instead of M and remember that to find moles, use the expression
(Molar)*(Volume) = moles
So we can write: (0.180 moles/liter)*(42.78 ml) = moles of NaOH
Note that we need a unit conversion for ml and liter: (1 liter/1000ml)
(0.180 moles/liter)*(42.78 ml)*(1 liter/1000ml) = moles of NaOH
Now both the liter and ml units cancel, leaving just moles, which is want we want.
moles of NaOH = 0.0077 moles of NaOH
Since we know from the balanced equation that we need the same number of moles of HBr to have complete titration (neutralization), we can say that HBr is also = 0.0077 moles.
Now we can calculate the concentration of the 19.32 ml of HBr. We have 0.0077 moles of HBr in 19.32 ml:
We can write (0.0077 moles HBr)/(19.32 ml). That would give us a concentration in moles/ml. If we want the more formal Molar, we can use a conversion factor:
[(0.0077 moles HBr)/(19.32 ml)]*(1000 ml/liter)
= 0.3986M [moles/liter]
The concentration of HBr is 0.399M to 3 sig figs
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Note:
A useful relationship when working with dilutions or titrations is:
M1V1 = M2V2 The product of Molarity and Volume for 1 (the initial concentration of the solution being titrated) is equal to the product of Molarity and Volume for 2 (the final concentration or the titrating solution).
In this case:
State 1 is for the HBr
M1 = Unknown
V1 = 19.32ml
State 2 is for the NaOH
M2 = 0.180M
V2 = 42.78ml
Calculation:
M1V1 = M2V2
M1*(19.32ml) = (0.180M)*(43.78ml)
M1 = [(0.180M)*(43.78ml)]/(19.32ml)
This is convenient since we can see the ml units both cancel, leaving only M, the desired unit. A conversion factor for ml and liters is not required.
M1 = 0.40M