63.45 ml of 0.250 m ba(oh)₂ reacts with 0.750 m hno₃ and reaches the endpoint using phenolphthalein as an…

63.45 ml of 0.250 m ba(oh)₂ reacts with 0.750 m hno₃ and reaches the endpoint using phenolphthalein as an indicator. what volume of hno₃ is required for this reaction?\n2hno₃ + ba(oh)₂ → ba(no₃)₂ + 2h₂o\n? ml hno₃

63.45 ml of 0.250 m ba(oh)₂ reacts with 0.750 m hno₃ and reaches the endpoint using phenolphthalein as an indicator. what volume of hno₃ is required for this reaction?\n2hno₃ + ba(oh)₂ → ba(no₃)₂ + 2h₂o\n? ml hno₃

Answer

Explanation:

Step1: Find moles of ( Ba(OH)_2 )

Molarity ( M = \frac{n}{V} ), so ( n = M \times V ). Convert volume to liters: ( 63.45 , mL = 0.06345 , L ).
( n_{Ba(OH)_2} = 0.250 , M \times 0.06345 , L = 0.0158625 , mol ).

Step2: Use stoichiometry to find moles of ( HNO_3 )

From the reaction: ( 2 , mol , HNO_3 ) reacts with ( 1 , mol , Ba(OH)2 ).
( n{HNO_3} = 2 \times n_{Ba(OH)_2} = 2 \times 0.0158625 , mol = 0.031725 , mol ).

Step3: Find volume of ( HNO_3 )

Rearrange ( M = \frac{n}{V} ) to ( V = \frac{n}{M} ).
( V_{HNO_3} = \frac{0.031725 , mol}{0.750 , M} = 0.0423 , L ). Convert to mL: ( 0.0423 , L \times 1000 = 42.3 , mL ).

Answer:

42.3