a 25.0ml dsmplr og 0.105M HCI was titrated with 31.5ml of NaOH. What is the concentration of the NaOH
I got 0.0834M am I correct
HCI + NaOH -----> H20 + NaCl
25.0ml HCI x 1L/1000mlx0.105 mol HCI/1L HCI X 1 mol NaOH/ I mol HCI = 0.0834M
My calculator reads 0.0833333 which I would round to 0.0833 and not 0.0834.
Yes, you are correct. To find the concentration of NaOH, you need to use the balanced chemical equation:
HCI + NaOH -----> H2O + NaCl
From the equation, you can see that 1 mole of HCI reacts with 1 mole of NaOH. So, the moles of NaOH can be calculated from the moles of HCI.
First, convert the volume of HCI to liters:
25.0 ml (1 L/1000 ml) = 0.025 L
Then, use the molarity of HCI to calculate the moles of HCI:
0.025 L × 0.105 mol HCI/1 L HCI = 0.002625 mol HCI
Since the ratio of HCI to NaOH is 1:1, the moles of NaOH will be the same as the moles of HCI:
0.002625 mol NaOH
Finally, convert the moles of NaOH to molarity using the volume of NaOH used in the titration:
0.002625 mol NaOH / 0.0315 L NaOH = 0.0834 M
Therefore, the concentration of NaOH is indeed 0.0834 M.