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You need to know 3 out of 4 variables to calculate the unknown.
Formula:
V1*M1=V2*M2
In your question only V1 (=50 mL) and M2 (=0.24m) are known.
Eg. V2 (volume of added NaOH in mL) has to be determined (most probably your buret readings; I hope you still can find it back in your lab journal)
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What is the molarity of an naoh solution if 4.37ml is titrated by 11.1 ml of 0.0904 m hno3?
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
50 ml of hcl is titrated with a solution of 0.24 m naoh it requires 35 ml of naoh to reach the equivalence point what is the concentration of the hcl solution?
.17
20 ml of acetic acid ch3cooh is titrated with a solution of 0.15 m naoh if 35 ml of naoh are required to reach the equivalence point what is the concentration of the acetic acid solution?
0.26
20 ml of acetic acid is titrated with a solution of 0.15 M NaOH If 35 ml of NaOH are required to reach the equivalence point what is the concentration of the acetic acid solution?
0.01 molar
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
What is the molarity of an naoh solution if 4.37ml is titrated by 11.1 ml of 0.0904 m hno3?
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
50 ml of hcl is titrated with a solution of 0.24 m naoh it requires 35 ml of naoh to reach the equivalence point what is the concentration of the hcl solution?
.17
20 ml of acetic acid is titrated with a solution of 0.15 M NaOH If 35 ml of NaOH are required to reach the equivalence point what is the concentration of the acetic acid solution?
0.01 molar
20 ml of acetic acid ch3cooh is titrated with a solution of 0.15 m naoh if 35 ml of naoh are required to reach the equivalence point what is the concentration of the acetic acid solution?
0.26
Explain how you would make 50 ml of 0.15 m NaOH solution from a 3.05 m NaOH solution?
Take 2.409 ml of 3.05M NaOH mix c a balance of H2O to form 50ml of mixture
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
20ml of acetic acid is titrated with a solutuion of 0.15 m naoh if 35 ml of naoh are required to reach the equivalence point what is the concentration of the acetic acid solution?
.26
A solution containing 3.76 g naoh in 361 ml water is titrated with 491 ml hcl what is the concentration of hcl in the original solution?
The concentration of hcl is 0.13.
A 0.361 molar solution of the weak acid HA with a pKa of 4.039 is titrated with a 0.163 molar solution of NaOH. What is the pH of the solution at the equivalence point of this titration?
You need to know the volume of the weak acid being titrated so you can find how many moles of base are needed to match that of the acid.
What is the chemical equation for the neutralization reaction of HCl titrated with NaOH?
HCl + NaOH -----> NaCl + H2O
Best indicator for lactic acid when titrated against standard NaOH?
phenolphthalein.
How is rm value checked for milk?
Procedure:- · 1-5g of sample is taken in a polenske flask. · 2-20g glycerol and 2ml conc. NaOH (50%) is added to it. · 3-It is then heated over flame until clear solution is obtained. · 4-It is allowed to cool and 90ml boiling distilled water is added. · 5-4-5 glass beads and 50ml dil. Sulphuric Acid is added to it. · 6-Distillation is carried out for 18-21 minutes and 110ml distillate is collected. · 7-It is then cooled for 10 minutes at 15C. · 8-It is titrated with N/10 NaOH. Formula used: - R.M Value= (A- B) x N x 11 or (A- B) x 1.1 Where A= Volume in ml of Standard NaOH solution. B= Volume in ml of Standard NaOH solution for blank. N= Normality of Standard NaOH solution
