Question

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. it has the unwieldy formula of khc8h4o4. this is often written in shorthand notation as khp. how many grams of khp are needed to exactly neutralize 28.5 ml of a 0.483 m barium hydroxide solution ?

Answer 1

$2 KHP(aq) + Ba(OH)_{2}(aq) ---> KP_{2}Ba (aq) + 2 H_{2}O (l)$\

Moles of $Ba(OH)_{2}$=$0.483 \frac{mol}{L} * 28.5 mL *\frac{1 L}{1000mL} = 0.0138 mol Ba(OH)_{2}$

Mass of KHP = $0.0138 mol Ba(OH)_{2} * \frac{2 mol KHP}{1 mol Ba(OH)_{2}} * \frac{204.22 g KHP}{1mol KHP}$

= 5.64 g KHP

Answer 2

Answer:

5.64g of KHP

Explanation:

2KHC8H4O4 + Ba (OH)2 → Ba (KC8H4O4)2 + 2H2O

Number of moles (n) = concentration × volume

Number of moles (n) = 0.483× 28.5/1000= 0.0138 moles of Ba(OH)2

From the reaction equation:

2 moles of KHP reacted with 1 moles of Ba(OH)2

x moles of KHP will react with 0.0138 moles of Ba(OH)2

x= 2× 0.0138/1 = 0.0276 moles

Molar mass of KHP = 204.22 g/mol

Therefore mass of KHP = 204.22 g/mol × 0.0276 moles= 5.64g of KHP