Respuesta :
the balanced equation for the above reaction is
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCk is 11:
number of HCl moles reacted - 0.105 mol/L x 0.0250 L = 0.00263 mol
according to molar ratio, number of HCl moles = number of NaOH moles
therefore number of NaOH moles - 0.00263 mol
number of moles in 31.5 mL - 0.00263 mol
therefore number of NaOH moles in 1000 mL = 0.00263 / 31,5 mL x 1000 mL
molarity of NaOH - 0.0835 M
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCk is 11:
number of HCl moles reacted - 0.105 mol/L x 0.0250 L = 0.00263 mol
according to molar ratio, number of HCl moles = number of NaOH moles
therefore number of NaOH moles - 0.00263 mol
number of moles in 31.5 mL - 0.00263 mol
therefore number of NaOH moles in 1000 mL = 0.00263 / 31,5 mL x 1000 mL
molarity of NaOH - 0.0835 M
The concentration of the NaOH: 0.083 M
Further explanation
Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint / equivalence point of the reaction can use indicators according to the appropriate pH range
Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.
An acid-base titration there will be a change in the pH of the solution.
From this pH change a Titration Curve can be made which is a function of acid / base volume and pH of the solution
Acid-base titration formula
Ma Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence
Neutralization Reaction:
NaOH + HCl ⇒ NaCl + H₂O
25.0 ml, 0.105 M HCl was titrated with 31.5 ml of NaOH
Acid-base titration formula
Ma Va. na = Mb. Vb. nb
a = HCl, b = NaOH (both have valence 1)
0.105 M. 25 = Mb. 31.5. 1
[tex]\rm Mb = \dfrac {0.105 \times 25} {31.5} \\\\ Mb = \boxed {\bold {0.083 \: M}}[/tex]
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