Question
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
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
Answers
My calculator reads 0.0833333 which I would round to 0.0833 and not 0.0834.
Related Questions
A sample of 1.029g of KC8H5O4 was titrated with an NAOH solution of unknown concentration. If 22.83...
A 5.0 M solution of HNO3 is titrated with 0.3 M NaOH. Identify the species that have the highest con...
A 25,0ml sample of 0,105M HCl was tritrated with 31,5ml of NaOH.what is the concentration of the NaO...
Find concentration acid in 50ml of 1.0 M NaOH is titrated with 75ml of sulfuric acid
2. 120cubic ce...