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11Alexandr11 [23.1K]
3 years ago
13

HELP!!! can anyone tell me if this virtual titration lab is right & help me find the missing concentrations ??

Chemistry
2 answers:
kow [346]3 years ago
6 0
The equation that you need to use is M₁V₁=M₂V₂ for both cases since the molar ratio is 1.

In titration 1:
You would use the equation M₂=M₁V₁/V₂

M₁=concentration of NaOH (0.1M)
V₁=volume of NaOH added (16mL)
V₂=volume of HCl in the Erlenmeyer flask (20mL) 
M₂=concentration of HCl in the Erlenmeyer flask (unknown)

M₂=(0.1M*16mL)/20mL
M₂=0.08M
The concentration of acid was 0.08M

Titration 2:
Still use the formula M₂=M₁V₁/V₂

M₁=concentration of HCL (0.1M)
V₁=volume of HCl added (14mL)
V₂=volume of NaOH in Erlenmeyer flask (20mL)
M₂=concentration of NaOH in the Erlenmeyer flask (unknown)

M₂=(0.1M*14mL)/20mL
M₂=0.07M
concentration of the NaOH is 0.07M

I hope this helps.  Let me know if you have any further questions or if anything needs to be clarified.
malfutka [58]3 years ago
3 0
Concentration is wrong, conc=moles/volume, both concentration are wrong use the formula to correct it.
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\large \boxed{\text{-92 $^{\circ}$C}}

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Data  

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Calculations

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\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

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