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vaieri [72.5K]
3 years ago
8

a quantity of 18.68 mL of KOH solution is needed to neutralize 0.4218 g of KHP. What is the concentration (in molarity) of the K

OH solution?
Chemistry
1 answer:
SashulF [63]3 years ago
5 0
To neutralize an acidic substance with the basic solution, the number of moles should be equal. First, we calculate for the number of moles of KHP.

molar mass of KHP = (1 mol K)(39.1 g K/1 mol K) + (1 mol H)(1 g H/1 mol H) + (1 mol P)(30.97 g P / 1 mol P) = 71.07 g KHP/ mol

number of moles KHP = (0.4218 g KHP) x (1 mol KHP / 71.07 g KHP)
        n = 0.0059 moles

Let M be the molarity of the solution and equating the number of moles,
     0.0059 moles = (18.68 mL)(1 L/1000 mL)(M)
            M = 0.32 moles/L

<em>ANSWER: 0.32 M</em>
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1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

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