Respuesta :
Answer:
152 mL is the volume of KOH required to reach the equivalence point.
Explanation:
[tex]HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)[/tex]
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2[/tex]
[tex]V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL[/tex]
152 mL is the volume of KOH required to reach the equivalence point.
Answer:
152 ml.
Explanation:
Given:
Volume of HNO3 = 90 ml
Molar concentration of HNO3 = 0.59 M
Molar concentration of KOH = 0.35 M
Equation of the reaction
KOH + HNO3 --> KNO3 + H2O
Number of moles of HNO3 = molar concentration × volume
= 0.59 × 0.09
= 0.0531 moles.
By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,
Number of moles of KOH = 0.0531 moles.
Volume = 0.0531 ÷ 0.350
= 0.152 l
= 152 ml.
