A50.0-ml sample of a 1.50 m naoh solution is titrated with a 2.90 m hcl solution. what will be the final volume (sample + titrant) of solution when the naoh has been completely neutralized by the hcl?

In order to proceed calculations for a titration process, it is important that we know the reaction involved. For this case, it would be:

HCl + NaOH = NaCl + H2O

To determine the final volume of the sample and the titrant added, we need to know how much of the titrant or the HCl solution was added into the sample. To do this, we use the given values in the problem and the reaction to relate the substances. We do as follows:

1.50 M NaOH ( .05 L solution ) ( 1 mol HCl / 1 mol NaOH ) = 0.075 mol HCl
0.075 mol HCl / 2.90 M HCl  = 0.0259 L HCl solution or 25.9 mL of the HCl solution was added

Therefore, the total volume of the sample and the titrant would be 50.0 + 25.90 = 75.90 mL.

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer